13. (15 points) In $\triangle A B C$, the internal angles $\angle A$, $\angle B$, and $\angle C$ have opposite sides of lengths $a$, $b$, and $c$ respectively. Given that $c=2$ and $\angle C=\frac{\pi}{3}$. Find
(1) the maximum perimeter of $\triangle A B C$;
(2) the area of $\triangle A B C$ when $2 \sin 2 A+\sin (2 B+C)=\sin C$.
$$
\begin{array}{l}
(1) \text{ By the cosine theorem and the given conditions, we have} \\
a^{2}+b^{2}-a b=4 \\
\Rightarrow(a+b)^{2}=4+3 a b \\
\quad \leqslant 4+3 \times \frac{(a+b)^{2}}{4} \\
\Rightarrow a+b \leqslant 4 .
\end{array}
$$

Therefore, the maximum perimeter of $\triangle A B C$ is 6, which is achieved when $\triangle A B C$ is an equilateral triangle.
$$
\begin{array}{l}
(2) \text{ From } 2 \sin 2 A+\sin (2 B+C)=\sin C \\
\Rightarrow \sin (B+A)+\sin (B-A)=4 \sin A \cdot \cos A \\
\Rightarrow \sin B \cdot \cos A=2 \sin A \cdot \cos A .
\end{array}
$$

When $\cos A=0$,
$$
\angle A=\frac{\pi}{2}, \angle B=\frac{\pi}{6}, a=\frac{4 \sqrt{3}}{3}, b=\frac{2 \sqrt{3}}{3} \text {. }
$$

When $\cos A \neq 0$, $\sin B=2 \sin A$.
By the sine theorem, we get $b=2 a$.
Solving the system of equations $\left\{\begin{array}{l}a^{2}+b^{2}-a b=4, \\ b=2 a .\end{array}\right.$
We get $a=\frac{2 \sqrt{3}}{3}, b=\frac{4 \sqrt{3}}{3}$.
Thus, $S_{\triangle A B C}=\frac{1}{2} a b \sin C=\frac{2 \sqrt{3}}{3}$.
$$