= Solution
In the basis $(|\psi_1\rangle,|\psi_2\rangle)$,
$$
H=\begin{pmatrix}E_1&h\\h&E_2\end{pmatrix}.
$$
Writing $\Delta=E_2-E_1>0$, its exact eigenvalues are
$$
\boxed{
E_\pm=\frac{E_1+E_2}{2}
\pm\sqrt{\frac{\Delta^2}{4}+h^2}.}
$$
For the normalized trial state,
$$
\begin{aligned}
E(\beta)
&=E_1\sin^2\beta+E_2\cos^2\beta
+2h\sin\beta\cos\beta\\
&=\frac{E_1+E_2}{2}
+\frac\Delta2\cos2\beta+h\sin2\beta.
\end{aligned}
$$
The minimum of $A\cos2\beta+h\sin2\beta$ is $-\sqrt{A^2+h^2}$ with $A=\Delta/2$. Therefore
$$
\boxed{
\min_\beta E(\beta)
=\frac{E_1+E_2}{2}
-\sqrt{\frac{\Delta^2}{4}+h^2}=E_-.}
$$
The variational estimate is exact because the trial family ranges over all real normalized states in the two-dimensional Hilbert space, and the real symmetric Hamiltonian has a real ground eigenvector.
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