For the Hull-White model, where
\begin{align*}
dr_t = (\theta(t)-a r_t)dt+ \sigma dW_t,
\end{align*}
under the risk-neutral measure, we have that, for $t\ge s \ge 0$,
\begin{align*}
r_t = e^{-a(t-s)} r_s + \int_s^t \theta(u)e^{-a(t-u)} du + \int_s^t \sigma e^{-a(t-u)} dW_u.
\end{align*}
Then, if $\theta$ is a constant,
\begin{align*}
r_t \mid r_s &\sim N\left(e^{-a(t-s)} r_s + \int_s^t \theta e^{-a(t-u)} du\Big), \, \frac{\sigma^2}{2a}\Big(1-e^{-2a(t-s)}\Big)\right) \\
&\sim N\left(e^{-a(t-s)} r_s + \frac{\theta}{a} \Big(1-e^{-a(t-s)}\Big), \, \frac{\sigma^2}{2a}\Big(1-e^{-2a(t-s)}\Big)\right).
\end{align*}
For the general case (see this question), the price of a zero-coupon bond price is given by
\begin{align*}
P(t, T) &= A(t, T) e^{-B(t, T)\, r_t},
\end{align*}
where
\begin{align*}
B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big),
\end{align*}
and
\begin{align*}
A(t, T) &= \exp\left(- \int_t^T \theta(u) B(u, T) du -\frac{\sigma^2}{2a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{4a}B(t, T)^2\right).
\end{align*}
Given the initial bond price curve, note that
\begin{align*}
\ln P(0, T) = \ln A(0, T) - B(0, T)\, r_0.
\end{align*}
Then
\begin{align*}
f(0, T) &= -\frac{\partial \ln P(0, T)}{\partial T}\\
&= \int_0^T \theta(u) \frac{\partial B(u, T)}{\partial T} du + \frac{\sigma^2}{2a^2}\Big(\frac{\partial B(0, T)}{\partial T} -1\Big)+ \frac{\sigma^2}{2a}B(0, T) \frac{\partial B(0, T)}{\partial T}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ + \frac{\partial B(0, T)}{\partial T} r_0\\
&=\int_0^T \theta(u) e^{-a(T-u)} du+ \frac{\sigma^2}{2a^2}\Big(e^{-a T} -1\Big)+ \frac{\sigma^2}{2a^2}\Big(e^{-a T}-e^{-2a T} \Big) + e^{-a T} r_0\\
&=\int_0^T \theta(u) e^{-a(T-u)} du - \frac{\sigma^2}{2a^2}\Big(e^{-a T} -1\Big)^2 + e^{-a T} r_0.
\end{align*}
That is,
\begin{align*}
\int_0^T \theta(u) e^{-a(T-u)} du &= f(0, T) + \frac{\sigma^2}{2a^2}\Big(e^{-a T} -1\Big)^2-e^{-a T} r_0\\
&=\alpha(T)-e^{-a T} r_0, \tag{1}
\end{align*}
where
\begin{align*}
\alpha(T) = f(0, T) + \frac{\sigma^2}{2a^2}\Big(e^{-a T} -1\Big)^2.
\end{align*}
Moreover, from $(1)$,
\begin{align*}
\int_0^T \theta(u) e^{au} du= e^{aT}\alpha(T)-r_0.
\end{align*}
Then
\begin{align*}
\int_s^t \theta(u) e^{-a(t-u)} du &= e^{-a t} \int_s^t \theta(u) e^{a u} du\\
&= e^{-a t}\left(e^{at}\alpha(t)-e^{as} \alpha(s) \right)\\
&=\alpha(t) - e^{-a(t-s)}\alpha(s).
\end{align*}
Therefore,
\begin{align*}
r_t \mid r_s &\sim N\left(e^{-a(t-s)} r_s + \int_s^t \theta(u) e^{-a(t-u)} du\Big), \, \frac{\sigma^2}{2a}\Big(1-e^{-2a(t-s)}\Big)\right) \\
&\sim N\left(e^{-a(t-s)} r_s + \alpha(t) - e^{-a(t-s)}\alpha(s), \, \frac{\sigma^2}{2a}\Big(1-e^{-2a(t-s)}\Big)\right).
\end{align*}