已知向量 $\vec{a}=(\cos \alpha, \sin \alpha), \vec{b}=(\cos \beta, \sin \beta),|\vec{a}-\vec{b}|=\frac{2 \sqrt{5}}{5}$.

(1) 求 $\cos (\alpha-\beta)$ 的值;

(2) 若 $-\frac{\pi}{2}<\beta<0<\alpha<\frac{\pi}{2}$, 且 $\sin \beta=-\frac{5}{13}$, 求 $\sin \alpha$ 的值.
(1) $\because|\vec{a}-\vec{b}|=\frac{2 \sqrt{5}}{5}, \therefore \vec{a}^{2}-2 \vec{a} \cdot \vec{b}+\vec{b}^{2}=\frac{4}{5}$.

又 $\because \vec{a}=(\cos \alpha, \sin \alpha), \because \vec{b}=(\cos \beta, \sin \beta)$ ，

$\therefore \vec{a}^{2}=\vec{b}^{2}=1, \vec{a} \cdot \vec{b}=\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)$,

$\therefore \cos (\alpha-\beta)=\frac{3}{5}$.

(2) $\because-\frac{\pi}{2}<\beta<0<\alpha<\frac{\pi}{2}, \therefore 0<\alpha-\beta<\pi$.

由 (1) 得 $\cos (\alpha-\beta)=\frac{3}{5}, \therefore \sin (\alpha-\beta)=\frac{4}{5}$,

又 $\because \sin \beta=-\frac{5}{13}, \therefore \cos \beta=\frac{12}{13}$,

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