17. (10 points) Given the function
$$
f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\cos ^{2} x-\frac{1}{2}(x \in \mathbf{R}) \text {. }
$$
(1) When $x \in\left[-\frac{\pi}{12}, \frac{5 \pi}{12}\right]$, find the minimum and maximum values of the function $f(x)$;
(2) Let the internal angles $\angle A, \angle B, \angle C$ of $\triangle A B C$ correspond to the sides $a, b, c$ respectively, and $c=\sqrt{3}, f(C)=0$. If the vectors $\boldsymbol{\alpha}=(1, \sin A), \boldsymbol{\beta}=(2, \sin B)$ are collinear, find the values of $a, b$.
$$
\begin{array}{l}
f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1+\cos 2 x}{2}-\frac{1}{2} \\
=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2} \cos 2 x-1 \\
=\sin \left(2 x-\frac{\pi}{6}\right)-1
\end{array}
$$
$$
\begin{array}{l}
\text { (1) From }-\frac{\pi}{12} \leqslant x \leqslant \frac{5 \pi}{12} \\
\Rightarrow-\frac{\pi}{3} \leqslant 2 x-\frac{\pi}{6} \leqslant \frac{2 \pi}{3} \\
\Rightarrow-\frac{\sqrt{3}}{2} \leqslant \sin \left(2 x-\frac{\pi}{6}\right) \leqslant 1 \\
\Rightarrow-1-\frac{\sqrt{3}}{2} \leqslant \sin \left(2 x-\frac{\pi}{6}\right)-1 \leqslant 0 .
\end{array}
$$

Thus, the minimum value of $f(x)$ is $-1-\frac{\sqrt{3}}{2}$, and the maximum value is 0.
(2) From $f(C)=\sin \left(2 C-\frac{\pi}{6}\right)-1=0$, we know
$$
\begin{array}{l}
\sin \left(2 C-\frac{\pi}{6}\right)=1 . \\
\text { Also, } 0<\angle C<\pi \\
\Rightarrow-\frac{\pi}{6}<2 \angle C-\frac{\pi}{6}<\frac{11 \pi}{6} \\
\Rightarrow 2 \angle C-\frac{\pi}{6}=\frac{\pi}{2} \Rightarrow \angle C=\frac{\pi}{3} .
\end{array}
$$

Since vectors $\boldsymbol{\alpha}$ and $\boldsymbol{\beta}$ are collinear, thus, $\sin B=2 \sin A$.
By the Law of Sines, we get
$$
b=2 a \text {. }
$$

By the Law of Cosines, we get
$$
\begin{array}{l}
c^{2}=a^{2}+b^{2}-2 a b \cos \frac{\pi}{3} \\
\Rightarrow a^{2}+b^{2}-a \dot{b}=3 .
\end{array}
$$

Solving equations (1) and (2), we get $a=1, b=2$.