7. 已知向量 $\boldsymbol{m}=\left(\cos \alpha-\frac{\sqrt{2}}{3},-1\right), \boldsymbol{n}=(\sin \alpha, 1), \boldsymbol{m}$ 与 $\boldsymbol{n}$ 为共线向量, 且 $\alpha \in\left[-\frac{\pi}{2}, 0\right]$.

(1)求 $\sin \alpha+\cos \alpha$ 的值;

(2)求 $\frac{\sin 2 \alpha}{\sin \alpha-\cos \alpha}$ 的值.
(1) $\because \boldsymbol{m}$ 与 $\boldsymbol{n}$ 为共线向量,

$\therefore\left(\cos \alpha-\frac{\sqrt{2}}{3}\right) \times 1-(-1) \times \sin \alpha=0$,

即 $\sin \alpha+\cos \alpha=\frac{\sqrt{2}}{3}$.

(2) 由(1)得 $1+\sin 2 \alpha=(\sin \alpha+\cos \alpha)^{2}=\frac{2}{9}$,

$\therefore \sin 2 \alpha=-\frac{7}{9}$.

$\because(\sin \alpha+\cos \alpha)^{2}+(\sin \alpha-\cos \alpha)^{2}=2$,

$\therefore(\sin \alpha-\cos \alpha)^{2}=2-\left(\frac{\sqrt{2}}{3}\right)^{2}=\frac{16}{9}$.

又 $\because \alpha \in\left[-\frac{\pi}{2}, 0\right]$,

$\therefore \sin \alpha-\cos <0, \sin \alpha-\cos \alpha=-\frac{4}{3}$.

因此, $\frac{\sin 2 \alpha}{\sin \alpha-\cos \alpha}=\frac{7}{12}$.