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We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$, for $t>s$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-\mu(t-s)-X_s \le y-\mu(t-s)-X_s \mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu(t-s)-X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\mu(t-s)}X_s + \theta\left(1-e^{-\mu(t-s)} \right)+\sigma\int_s^te^{-\mu(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\mu(t-v)}dW_v \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\mu(t-s)}x - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*}

We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$, for $t>s$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-\mu(t-s)-X_s \le y-\mu(t-s)-X_s \mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu(t-s)-X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\theta(t-s)}X_s + \mu\left(1-e^{-\theta(t-s)} \right)+\sigma\int_s^te^{-\theta(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \le y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\theta(t-v)}dW_v \le y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\theta(t-s)}X_s - \mu\big(1-e^{-\theta(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\theta}\big(1-e^{-2\theta(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\theta(t-s)}x - \mu\big(1-e^{-\theta(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\theta}\big(1-e^{-2\theta(t-s)} \big)}} \right). \end{align*}

We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-X_s \le y-X_s \mid X_s)\\ &=P(\mu(t-s)+\sigma(W_t-W_s) \le y-X_s\mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu (t-s) -X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\mu(t-s)}X_s + \theta\left(1-e^{-\mu(t-s)} \right)+\sigma\int_s^te^{-\mu(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\mu(t-v)}dW_v \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\mu(t-s)}x - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*}

We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$, for $t>s$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-\mu(t-s)-X_s \le y-\mu(t-s)-X_s \mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu(t-s)-X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\mu(t-s)}X_s + \theta\left(1-e^{-\mu(t-s)} \right)+\sigma\int_s^te^{-\mu(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\mu(t-v)}dW_v \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\mu(t-s)}x - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*}

The idea for both processes are similar. For ease of exposition, we consider the simpler one, that is, $X_t = X_0 + \mu t + \sigma W_t$. Then, for $t>s$, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-X_s \le y-X_s \mid X_s)\\ &=P(\mu(t-s)+\sigma(W_t-W_s) \le y-\mu s -\sigma W_s\mid W_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu t -\sigma W_s\mid W_s)\\ &=\Phi\left(\frac{y-\mu t -\sigma W_s}{\sigma\sqrt{t-s}}\right)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the other one, solve $X_t$ first, and then do the similar computation.

We consider the first one, that is, $X_t = X_s + \mu (t-s) + \sigma (W_t-W_s)$. Then, \begin{align*} P(X_t \le y \mid X_s) &= P(X_t-X_s \le y-X_s \mid X_s)\\ &=P(\mu(t-s)+\sigma(W_t-W_s) \le y-X_s\mid X_s)\\ &=P(\sigma(W_t-W_s) \le y-\mu (t-s) -X_s\mid X_s)\\ &=\Phi\left(\frac{y-\mu (t-s) -X_s}{\sigma\sqrt{t-s}}\right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-\mu (t-s) -x}{\sigma\sqrt{t-s}}\right). \end{align*} Here, $\Phi$ is the cumulative distribution function of a standard normal random variable. The transition density function can be obtained subsequently by taking the derivative with respect to $y$.

For the second one, note that, for $t>s$, \begin{align*} X_t = e^{-\mu(t-s)}X_s + \theta\left(1-e^{-\mu(t-s)} \right)+\sigma\int_s^te^{-\mu(t-v)}dW_v. \end{align*} Then, \begin{align*} &\ P(X_t \le y \mid X_s)\\ =&\ P\left(X_t-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ P\left(\sigma\int_s^te^{-\mu(t-v)}dW_v \le y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big) \mid X_s\right)\\ =&\ \Phi\left(\frac{y-e^{-\mu(t-s)}X_s - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*} That is, \begin{align*} P(X_t \le y \mid X_s=x) &=\Phi\left(\frac{y-e^{-\mu(t-s)}x - \theta\big(1-e^{-\mu(t-s)} \big)}{\sigma\sqrt{\frac{1}{2\mu}\big(1-e^{-2\mu(t-s)} \big)}} \right). \end{align*}