13. (16 points) In an acute triangle $\triangle ABC$, $\sin A=\frac{4 \sqrt{5}}{9}$, and $a$, $b$, $c$ are the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively.
(1) Find the value of $\sin 2(B+C)+\sin ^{2} \frac{B+C}{2}$;
(2) If $a=4$, try to find the value of the area $S$ of $\triangle ABC$ when $\overrightarrow{A B} \cdot \overrightarrow{A C}$ is maximized.
13. (1) From the problem, we know $\cos A=\frac{1}{9}$.
Hence $\sin 2(B+C)+\sin ^{2} \frac{B+C}{2}$
$=\sin (2 \pi-2 A)+\frac{1-\cos (B+C)}{2}$
$=-2 \sin A \cdot \cos A+\frac{1+\cos A}{2}$
$=-2 \times \frac{4 \sqrt{5}}{9} \times \frac{1}{9}+\frac{1+\frac{1}{9}}{2}=\frac{45-8 \sqrt{5}}{81}$.
(2) It is easy to know, $\overrightarrow{A B} \cdot \overrightarrow{A C}=b c \cos A=\frac{1}{9} b c$.

By the cosine rule, we get
$$
\begin{array}{l}
16=a^{2}=b^{2}+c^{2}-2 b c \cos A \\
=b^{2}+c^{2}-\frac{2}{9} b c \geqslant 2 b c-\frac{2}{9} b c=\frac{16}{9} b c \\
\Rightarrow b c \leqslant 9 .
\end{array}
$$

When and only when $b=c=3$, all the above equalities hold.
Thus $S=\frac{1}{2} b c \sin A \leqslant \frac{1}{2} \times 9 \times \frac{4 \sqrt{5}}{9}=2 \sqrt{5}$.