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*}