(12 分) 已知 $\omega>0, \boldsymbol{a}=(\sqrt{3} \sin \omega x,-\cos \omega x), \boldsymbol{b}=(\cos \omega x, \cos \omega x)$, $f(x)=\boldsymbol{a} \cdot \boldsymbol{b}, x_{1}, x_{2}$ 是 $y=f(x)-\frac{1}{2}$ 的其中两个零点, 且 $\left|x_{1}-x_{2}\right|_{\min }=\pi$.

(1) 求 $f(x)$ 的单调递增区间;

(2) 若 $\alpha \in\left(0, \frac{\pi}{2}\right), f\left(\frac{\alpha}{2}\right)=\frac{1}{10}$, 求 $\sin 2 \alpha$ 的值.

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(1) $f(x)=\sqrt{3} \sin \omega x \cos \omega x-\cos ^{2} \omega x=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1+\cos 2 \omega x}{2}$

$=\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1}{2} \cos 2 \omega x-\frac{1}{2}=\sin \left(2 \omega x-\frac{\pi}{6}\right)-\frac{1}{2}$,

$\because x_{1}, x_{2}$ 是函数 $y=f(x)-\frac{1}{2}=\sin \left(2 \omega x-\frac{\pi}{6}\right)-1$ 的两个零点,

即 $x_{1}, x_{2}$ 是方程 $\sin \left(2 \omega x-\frac{\pi}{6}\right)=1$ 的两个实根, 且 $\left|x_{1}-x_{2}\right|_{\min }=\pi$,

$\therefore T=\pi, \quad \therefore 2 \omega=\frac{2 \pi}{\pi}=2$, 则 $\omega=1$,

$f(x)=\sin \left(2 x-\frac{\pi}{6}\right)-\frac{1}{2}$

令 $-\frac{\pi}{2}+2 k \pi \leq 2 x-\frac{\pi}{6} \leq \frac{\pi}{2}+2 k \pi, k \in \mathbf{Z}$, 得 $-\frac{\pi}{6}+k \pi \leq x \leq \frac{\pi}{3}+k \pi, k \in \mathbf{Z}$,

$\therefore f(x)$ 的单调递增区间为 $\left[-\frac{\pi}{6}+k \pi, \frac{\pi}{3}+k \pi\right](k \in \mathbf{Z})$.

(2) $f\left(\frac{\alpha}{2}\right)=\sin \left(\alpha-\frac{\pi}{6}\right)-\frac{1}{2}=\frac{1}{10}, \therefore \sin \left(\alpha-\frac{\pi}{6}\right)=\frac{3}{5}$.

$\because 0<\alpha<\frac{\pi}{2}, \therefore-\frac{\pi}{6}<\alpha-\frac{\pi}{6}<\frac{\pi}{3}, \therefore \cos \left(\alpha-\frac{\pi}{6}\right)=\frac{4}{5}$,

$\because \sin \alpha=\sin \left[\left(\alpha-\frac{\pi}{6}\right)+\frac{\pi}{6}\right]=\sin \left(\alpha-\frac{\pi}{6}\right) \cos \frac{\pi}{6}+\cos \left(\alpha-\frac{\pi}{6}\right) \sin \frac{\pi}{6}$ $=\frac{4+3 \sqrt{3}}{10}$

$\cos \alpha=\cos \left[\left(\alpha-\frac{\pi}{6}\right)+\frac{\pi}{6}\right]=\cos \left(\alpha-\frac{\pi}{6}\right) \cos \frac{\pi}{6}-\sin \left(\alpha-\frac{\pi}{6}\right) \sin \frac{\pi}{6}=\frac{4 \sqrt{3}-3}{10}$,

$\therefore \sin 2 \alpha=2 \sin \alpha \cos \alpha=2 \times \frac{4+3 \sqrt{3}}{10} \times \frac{4 \sqrt{3}-3}{10}=\frac{24+7 \sqrt{3}}{50}$.