13. In the acute triangle $\triangle ABC$, $\sin A=\frac{4 \sqrt{5}}{9}$, and $a, b, c$ are the sides opposite to angles $A, B, C$ respectively.
(1) Find the value of $\sin 2(B+C)+\sin ^{2} \frac{B+C}{2}$;
(2) If $a=4$, try to find the area $S_{\triangle A B C}$ when $\overrightarrow{A B} \cdot \overrightarrow{A C}$ is maximized.
$$
\begin{array}{l}
\text { (1) } \sin 2\left(180^{\circ}-A\right)+\sin ^{2}\left(90^{\circ}-\frac{A}{2}\right)=-\sin 2 A+\cos ^{2} \frac{A}{2} \\
=-2 \sin A \cos A+\frac{1}{2}(\cos A+1)=-2 \cdot \frac{4 \sqrt{5}}{9} \cdot \frac{1}{9}+\frac{1}{2} \cdot \frac{10}{9}=\frac{45-8 \sqrt{5}}{81} \text {. }
\end{array}
$$
(2) $a^{2}=16=b^{2}+c^{2}-\frac{2}{9} b c \geqslant \frac{16}{9} b c \Rightarrow b c \leqslant 9$, equality holds when $b=c=3$.

At this time, $\overrightarrow{A B} \cdot \overrightarrow{A C}=\frac{2}{9} b c$ reaches its maximum value, so $S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{9}{2} \cdot \frac{4 \sqrt{5}}{9}=2 \sqrt{5}$.