13. (15 points) Given an ellipse $E: \frac{x^{2}}{8}+\frac{y^{2}}{b^{2}}=1$ with its foci on the $x$-axis, which contains a circle $C: x^{2}+y^{2}=\frac{8}{3}$. A tangent line $l$ of circle $C$ intersects ellipse $E$ at points $A$ and $B$, and satisfies $\overrightarrow{O A} \perp \overrightarrow{O B}$ (where $O$ is the origin). Find
(1) the value of $b^{2}$;
(2) the range of $|A B|$.
13. (1) As shown in Figure 2, let the tangent line $l$ of circle $C$ intersect the ellipse $E$ at two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
When the slope of line $l$ exists, let the equation of $l$ be
$$
y=k x+m \text {. }
$$

Substituting into the equation of the ellipse, we get
$$
\left(b^{2}+8 k^{2}\right) x^{2}+16 m k x+8\left(m^{2}-b^{2}\right)=0 \text {. }
$$

By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{-16 m k}{b^{2}+8 k^{2}}, x_{1} x_{2}=\frac{8\left(m^{2}-b^{2}\right)}{b^{2}+8 k^{2}} \text {. }
$$

Since $\overrightarrow{O A} \perp \overrightarrow{O B}$, then
$$
x_{1} x_{2}+y_{1} y_{2}=0 \text {. }
$$

Substituting equation (1) into equation (3), we get
$$
\left(1+k^{2}\right) x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}=0 \text {. }
$$

Substituting equation (2) into the above equation, we get
$$
m^{2}=\frac{8 b^{2}\left(1+k^{2}\right)}{8+b^{2}} \text {. }
$$

Since line $l$ is tangent to the circle, the distance from the center of the circle to the tangent line equals the radius, i.e., $R=\frac{|m|}{\sqrt{1+k^{2}}}$.
Thus, $m^{2}=\frac{8}{3}\left(1+k^{2}\right)$.
Substituting into equation (4), we solve to get $b^{2}=4$.
When the slope of line $l$ does not exist, let the equation of $l$ be $x=\sqrt{\frac{8}{3}}$.
Substituting into the equation of the ellipse, we get $y= \pm \sqrt{\frac{2}{3}} b$.
By equation (3), we have
$$
\frac{8}{3}-\frac{2}{3} b^{2}=0 \Rightarrow b^{2}=4 \text {. }
$$

In summary, regardless of whether the slope of line $l$ exists, we always have $b^{2}=4$.
(2) Draw $O D \perp A B$ at point $D$. Then $D$ is the point of tangency.

Thus, $O D=R=\frac{2 \sqrt{6}}{3}$.
Let $\angle O A B=\theta$. Then $\theta$ is an acute angle, and
$|A D|=\frac{R}{\tan \theta}=\frac{2 \sqrt{6}}{3 \tan \theta}$,
$|B D|=\frac{2 \sqrt{6}}{3} \tan \theta$.
Therefore, $|A B|=|A D|+|B D|$
$=\frac{2 \sqrt{6}}{3}\left(\tan \theta+\frac{1}{\tan \theta}\right)$.
Since $|O A|$ varies between the semi-major and semi-minor axes of the ellipse, i.e., $2 \leqslant|O A| \leqslant 2 \sqrt{2}$,
we have $\frac{\sqrt{2}}{2} \leqslant \tan \theta \leqslant \sqrt{2}$.
Let $x=\tan \theta$. It is easy to prove that when $x \in\left[\frac{\sqrt{2}}{2}, 1\right]$,
$$
|A B|=\frac{2 \sqrt{6}}{3}\left(x+\frac{1}{x}\right)
$$

is monotonically decreasing;
when $x \in[1, \sqrt{2}]$,
$$
|A B|=\frac{2 \sqrt{6}}{3}\left(x+\frac{1}{x}\right)
$$

is monotonically increasing.
Therefore, when $x=1$, $|A B|_{\min }=\frac{4 \sqrt{6}}{3}$;
when $x=\sqrt{2}$ or $\frac{\sqrt{2}}{2}$, $|A B|_{\text {max }}=2 \sqrt{3}$.
In summary, $\frac{4 \sqrt{6}}{3} \leqslant|A B| \leqslant 2 \sqrt{3}$.