13. Through the right focus $F$ of the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, draw two perpendicular chords $A B, C D$. Let the midpoints of $A B, C D$ be $M, N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If the slopes of $A B, C D$ both exist, find the maximum value of the area of $\triangle F M N$.
(1) As shown in the figure, let the line $AB: x=ny+1$, and combine
$$
\left\{\begin{array}{l}
x=ny+1, \\
4x^{2}+5y^{2}=20
\end{array} \Rightarrow \left(4n^{2}+5\right)y^{2}+8ny-16=0\right.
$$
$\Rightarrow M\left(\frac{5}{4n^{2}+5},-\frac{4n}{4n^{2}+5}\right)$, substituting $-\frac{1}{n}$ for $n$, we get
$$
N\left(\frac{5n^{2}}{5n^{2}+4}, \frac{4n}{5n^{2}+4}\right) \Rightarrow k_{MN}=\frac{9n}{5\left(n^{2}-1\right)} \text{. }
$$

Thus, the line $MN: y+\frac{4n}{4n^{2}+5}=\frac{9n}{5\left(n^{2}-1\right)}\left(x-\frac{5}{4n^{2}+5}\right)$, setting $y=0$ gives $x=\frac{4}{9} \cdot \frac{5\left(n^{2}-1\right)}{4n^{2}+5}+\frac{5}{4n^{2}+5}=\frac{5}{9}$, so the line $MN$ must pass through the fixed point $E\left(\frac{5}{9}, 0\right)$.
(2) From (1), we know $S_{FMN}=\frac{1}{2}|EF| \cdot\left|y_{M}-y_{N}\right|=\frac{2}{9}\left|\frac{4n}{4n^{2}+5}+\frac{4n}{5n^{2}+4}\right|$
$$
=\frac{8|n|\left(n^{2}+1\right)}{\left(4n^{2}+5\right)\left(5n^{2}+4\right)}=\frac{8|n|\left(n^{2}+1\right)}{n^{2}+20\left(n^{2}+1\right)^{2}}=\frac{8\left|n+\frac{1}{n}\right|}{1+20\left(n+\frac{1}{n}\right)^{2}} \xlongequal{t=\left|n+\frac{1}{n}\right| \geqslant 2} \frac{8}{\frac{1}{t}+20t} \leqslant \frac{16}{81} \text{, }
$$

The equality holds when $t=2 \Rightarrow n= \pm 1$. In summary, the maximum area of $\triangle FMN$ is $\frac{16}{81}$.