1. Let $\angle A, \angle B, \angle C$ be the three interior angles of $\triangle ABC$, and the complex number
$$
\begin{array}{l}
z=\frac{\sqrt{65}}{5} \sin \frac{A+B}{2}+\mathrm{i} \cos \frac{A-B}{2}, \\
|z|=\frac{3 \sqrt{5}}{5} .
\end{array}
$$

Then the maximum value of $\angle C$ is $\qquad$
$=1 . \pi-\arctan \frac{12}{5}$. 
From $|z|^{2}=\frac{9}{5}$, we know
$$
\frac{13}{5} \sin ^{2} \frac{A+B}{2}+\cos ^{2} \frac{A-B}{2}=\frac{9}{5} \text {, }
$$

we get $13 \cos (A+B)=5 \cos (A-B)$.
Further, we have $\tan A \cdot \tan B=\frac{4}{9}$.
Thus, $\angle A$ and $\angle B$ are both acute angles, and
$$
\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \geqslant \frac{12}{5} \text {. }
$$

Therefore, $\angle A+\angle B \geqslant \arctan \frac{12}{5}$.
Hence $\angle C \leqslant \pi-\arctan \frac{12}{5}$.
It is easy to verify that equality can be achieved.
Thus, the maximum value sought is $\pi-\arctan \frac{12}{5}$.