1. From $\cos A= \pm \sqrt{1-\sin ^{2} A}$
$$
\begin{aligned}
& = \pm \sqrt{1-\left(\frac{3}{5}\right)^{2}}= \pm \frac{4}{5} \quad(A \in(0, \pi)) . \\
\sin B & =\sqrt{1-\cos ^{2} B} \\
& =\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13} \quad\left(B \in\left(0, \frac{\pi}{2}\right)\right) .
\end{aligned}
$$

If $\cos A=-\frac{4}{5}$, then
$$
\begin{aligned}
\sin C & =\sin [\pi-(A+B)]=\sin (A+B) \\
& =\sin A \cos B+\cos A \sin B \\
& =\frac{3}{5} \times \frac{5}{13}+\left(-\frac{4}{5}\right) \times \frac{12}{13}<0$, which is a contradiction.
$\therefore \cos A \neq-\frac{4}{5}$, so $\cos A=\frac{4}{5}$.
At this point, $\cos C=\cos [\pi-(A+B)]$
$$
\begin{array}{l}
=\sin A \sin B-\cos A \cos B \\
=\frac{3}{5} \times \frac{12}{13}-\frac{4}{5} \times \frac{5}{13}=\frac{16}{65} .
\end{array}
$$
1. In $\triangle A B C$, $\sin A=\frac{3}{5}, \cos B=\frac{5}{13}$, then the value of $\cos C$ is $\qquad$.