已知 $\sin \alpha=\frac{5}{13}$, 且 $0<\alpha<\frac{\pi}{2}$.

(1)求 $\sin 2 \alpha$ 的值;

(2)若 $\cos (\alpha-\beta)=\frac{4}{5}, 0<\alpha<\beta<\frac{\pi}{2}$, 求 $\cos \beta$ 的值.
(1) $\because 0<\alpha<\frac{\pi}{2}, \sin \alpha=\frac{5}{13}, \therefore \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{12}{13}$ ，

$\therefore \sin 2 \alpha=2 \sin \alpha \cos \alpha=\frac{120}{169}$.

(2) 若 $\cos (\alpha-\beta)=\frac{4}{5}, \quad 0<\alpha<\beta<\frac{\pi}{2}$, 则 $-\frac{\pi}{2}<\alpha-\beta<0$,
$\therefore \sin (\alpha-\beta)=-\sqrt{1-\cos ^{2}(\alpha-\beta)}=-\frac{3}{5}$,

$\therefore \cos \beta=\cos [\alpha-(\alpha-\beta)]=\cos \alpha \cos (\alpha-\beta)+\sin \alpha \sin (\alpha-\beta)=\frac{12}{13} \cdot \frac{4}{5}+\frac{5}{13} \cdot\left(-\frac{3}{5}\right)=\frac{33}{65}$.