13. (20 points) Given the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, a line $l$ that does not pass through the origin intersects the ellipse at points $A$ and $B$.
(1) Find the maximum value of the area of $\triangle O A B$.
(2) Does there exist an ellipse $C_{2}$ such that for every tangent line of $C_{2}$ intersecting the ellipse $C_{1}$ at points $A$ and $B$, the area $S_{\triangle O A B}$ is exactly the maximum value? If it exists, find the equation of the ellipse; if not, explain the reason.
13. (1) If the slope of line $l$ exists, let the equation of $l$ be $y=k x+m$, substituting into the ellipse equation yields
$\left(a^{2} k^{2}+b^{2}\right) x^{2}+2 a^{2} k m x+a^{2} m^{2}-a^{2} b^{2}=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$x_{1}+x_{2}=-\frac{2 a^{2} k m}{a^{2} k^{2}+b^{2}}, x_{1} x_{2}=\frac{a^{2} m^{2}-a^{2} b^{2}}{a^{2} k^{2}+b^{2}}$.
Thus, $|A B|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{1+k^{2}} \sqrt{\left(x_{2}+x_{1}\right)^{2}-4 x_{1} x_{2}}$
$=\sqrt{1+k^{2}} \frac{2 a b}{a^{2} k^{2}+b^{2}} \sqrt{a^{2} k^{2}+b^{2}-m^{2}}$.
In $\triangle O A B$, let the height from $O$ to $A B$ be $h$. Then
$h=\frac{|m|}{\sqrt{1+k^{2}}}$,
$S_{\triangle O A B}=\frac{1}{2}|A B| h=\frac{a b|m|}{a^{2} k^{2}+b^{2}} \sqrt{a^{2} k^{2}+b^{2}-m^{2}}$.
Fixing $k$, we have
$S_{\triangle O A B}=\frac{a b|m|}{a^{2} k^{2}+b^{2}} \sqrt{a^{2} k^{2}+b^{2}-m^{2}} \leqslant \frac{a b}{2}$.
Thus, for any $k$, $S_{\triangle O A B} \leqslant \frac{a b}{2}$, with equality if and only if $a^{2} k^{2}+b^{2}=2 m^{2}$.
If the slope of line $l$ does not exist, let the line $l: x=m$.
Then it is easy to prove that $S_{\triangle O A B} \leqslant \frac{a b}{2}$, with equality if and only if $a^{2}=2 m^{2}$.
In summary, the maximum area of $\triangle O A B$ is $\frac{a b}{2}$.
(2) There exists an ellipse $C_{2}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{2}$, such that any tangent line to this ellipse intersects the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at points $A$ and $B$, and $S_{\triangle O A B}=\frac{a b}{2}$.

In fact, let the ellipse satisfying the condition be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\lambda$. The equation of the tangent line at any point $\left(x_{0}, y_{0}\right)$ on the ellipse is
$$
\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=\lambda,
$$

This tangent line intersects the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at points $A$ and $B$.
If $y_{0}=0$, then $x_{0}= \pm a \sqrt{\lambda}$.
From the tangent line equation, we get $x=\frac{\lambda a^{2}}{x_{0}}= \pm a \sqrt{\lambda}$.
From (1), we know that $S_{\triangle O A B}=\frac{a b}{2}$ if and only if
$$
a^{2}=2(\sqrt{\lambda} a)^{2}=2 \lambda a^{2} \Leftrightarrow \lambda=\frac{1}{2} \text {. }
$$

Next, we prove that if $y_{0} \neq 0$, when $\lambda=\frac{1}{2}$, $S_{\triangle O A B}=\frac{a b}{2}$ still holds.

In this case, the tangent line equation at any point $\left(x_{0}, y_{0}\right)$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{2}$ is
$$
y=-\frac{b^{2}}{y_{0}} \cdot \frac{x_{0}}{a^{2}} x+\frac{b^{2}}{2 y_{0}} .
$$

Let $k=-\frac{b^{2}}{y_{0}} \cdot \frac{x_{0}}{a^{2}}, m=\frac{b^{2}}{2 y_{0}}$. Then
$a^{2} k^{2}+b^{2}=a^{2}\left(-\frac{b^{2}}{y_{0}} \cdot \frac{x_{0}}{a^{2}}\right)^{2}+b^{2}$
$=b^{2}\left(1+\frac{b^{2}}{y_{0}^{2}} \cdot \frac{x_{0}^{2}}{a^{2}}\right)$.
Also, $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=\frac{1}{2}$, thus,
$a^{2} k^{2}+b^{2}=a^{2}\left(-\frac{b^{2}}{y_{0}} \cdot \frac{x_{0}}{a^{2}}\right)^{2}+b^{2}$
$=b^{2}\left(1+\frac{b^{2}}{y_{0}^{2}} \cdot \frac{x_{0}^{2}}{a^{2}}\right)=\frac{b^{4}}{2 y_{0}^{2}}=2 m^{2}$.
From (1), we get $S_{\triangle O A B}=\frac{a b}{2}$.
In summary, there exists an ellipse $C_{2}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{2}$, such that for every tangent line to ellipse $C_{2}$ intersecting ellipse $C_{1}$ at points $A$ and $B$, $S_{\triangle O A B}$ exactly takes the maximum value.