1. In $\triangle A B C$,
$$
\begin{array}{l}
z=\frac{\sqrt{65}}{5} \sin \frac{A+B}{2}+\mathrm{i} \cos \frac{A-B}{2}, \\
|z|=\frac{3 \sqrt{5}}{5} .
\end{array}
$$

Then the maximum value of $\angle C$ is
$-1 . \pi-\arctan \frac{12}{5}$.
From $|z|^{2}=\frac{13}{5} \sin ^{2} \frac{A+B}{2}+\cos ^{2} \frac{A-B}{2}=\frac{9}{5}$
$\Rightarrow 13 \cos (A+B)=5 \cos (A-B)$
$\Rightarrow \tan A \cdot \tan B=\frac{4}{9}$
$\Rightarrow \angle A 、 \angle B$ are both acute angles, and
$$
\begin{aligned}
& \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \geqslant \frac{12}{5} \\
\Rightarrow & \angle A+\angle B \geqslant \arctan \frac{12}{5} \\
\Rightarrow & \angle C \leqslant \pi-\arctan \frac{12}{5} .
\end{aligned}
$$

It is easy to verify that the equality can be achieved.