(17) (12 points) In $\triangle ABC$, the opposite sides of the internal angles $A, B, C$ are $a, b, c$ respectively. The vectors $m=(2 \sin B, -\sqrt{3})$, $n=\left(\cos 2B, 2 \cos ^{2} \frac{B}{2}-1\right)$, and $\boldsymbol{m} / / \boldsymbol{n}$.
(1) Find the size of the acute angle $B$;
(2) If $b=2$, find the maximum value of $S_{\triangle ABC}$.
17 (1) Because $\boldsymbol{m} / / \boldsymbol{n}$, so
$$
2 \sin B\left(2 \cos ^{2} \frac{B}{2}-1\right)=-\sqrt{3} \cos 2 B,
$$

so $\sin 2 B=-\sqrt{3} \cos 2 B$, i.e., $\tan 2 B=-\sqrt{3}$.
Since $B$ is an acute angle, $2 B \in(0, \pi), 2 B=\frac{2 \pi}{3}$, so $B=\frac{\pi}{3}$.
(2) Because $B=\frac{\pi}{3}, b=2$, by the cosine rule $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$, we get $a^{2}+c^{2}-a c-4=0$, and since $a^{2}+c^{2} \geqslant 2 a c$, substituting this in, we get $a c \leqslant 4$ (with equality if and only if $a=c=2$).
$S_{\triangle A B C}=\frac{1}{2} a c \sin B=\frac{\sqrt{3}}{4} a c \leqslant \sqrt{3}$ (with equality if and only if $a=c=2$).
Therefore, the maximum value of $S_{\triangle A B C}$ is $\sqrt{3}$.