17. (10 points) In $\triangle A B C$, the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. Let
$$
\begin{array}{l}
\boldsymbol{m}=(2 \sin B, -\sqrt{3}), \\
\boldsymbol{n}=\left(\cos 2 B, 2 \cos ^{2} \frac{B}{2}-1\right), \text{ and } \boldsymbol{m} / / \boldsymbol{n} .
\end{array}
$$
(1) Find the size of the acute angle $\angle B$;
(2) If $b=2$, find the maximum value of $S_{\triangle A B C}$.
And $a^{2}+c^{2} \geqslant 2 a c$, substituting into the above equation we get $a c \leqslant 4$.
Equality holds if and only if $a=c=2$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} a c \sin B=\frac{\sqrt{3}}{4} a c \leqslant \sqrt{3}$, and equality holds in equation (1) if and only if $a=c=2$.
Three, 17. (1) From $m / / n$, we know
$$
\begin{array}{l}
2 \sin B^{\left(2 \mathrm{~mm} \frac{\pi}{2}-1\right)}=-\sqrt{3} \cos 2 B \\
\Rightarrow \sin 2 B=-\sqrt{3} \cos 2 B \\
\Rightarrow \tan 2 B=-\sqrt{3} \\
\Rightarrow 2 \angle B=\frac{2 \pi}{3} \Rightarrow \angle B=\frac{\pi}{3} .
\end{array}
$$
(2) From $\angle B=\frac{\pi}{3}, b=2$ and the cosine rule, we know
$$
\begin{array}{l}
\Rightarrow \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \\
\Rightarrow a^{2}+c^{2}-a c-4=0 .
\end{array}
$$