已知 $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, 且 $\sin \alpha=\frac{1}{3}$.

(1) 求 $\sin 2 \alpha$ 的值;

(2) 若 $\sin (\alpha+\beta)=-\frac{3}{5}, \beta \in\left(0, \frac{\pi}{2}\right)$, 求 $\sin \beta$ 的值.
(I) $\because \alpha \in\left(\frac{\pi}{2}, \pi\right)$, 且 $\sin \alpha=\frac{1}{3}, \therefore \cos \alpha=-\frac{2 \sqrt{2}}{3}$,

于是 $\sin 2 \alpha=2 \sin \alpha \cos \alpha=-\frac{4 \sqrt{2}}{9}$;

(II) $\because \alpha \in\left(\frac{\pi}{2}, \pi\right), \beta \in\left(0, \frac{\pi}{2}\right), \therefore \alpha+\beta \in\left(\frac{\pi}{2}, \frac{3}{2} \pi\right)$, 结合 $\sin (\alpha+\beta)=-\frac{3}{5}$ 得: $\cos (\alpha+\beta)=-\frac{4}{5}$,于是

$\sin \beta=\sin [(\alpha+\beta)-\alpha]=\sin (\alpha+\beta) \cos \alpha-\cos (\alpha+\beta) \sin \alpha$

$=-\frac{3}{5} \cdot\left(-\frac{2 \sqrt{2}}{3}\right)-\left(-\frac{4}{5}\right) \cdot \frac{1}{3}=\frac{4+6 \sqrt{2}}{15}$.