(13) 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 $B$, satisfying $\overrightarrow{O A} \perp \overrightarrow{O B}$ (where $O$ is the origin).
(1) Find the value of $b^{2}$;
(2) Find the range of $|A B|$.
13. (1) Let the tangent line $l$ of circle $C$ intersect 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
$$
y=k x+m,
$$

Substituting into $\frac{x^{2}}{8}+\frac{y^{2}}{b^{2}}=1$, we get
$$
\left(b^{2}+8 k^{2}\right) x^{2}+16 m k x+8\left(m^{2}-b^{2}\right)=0 .
$$

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

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

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

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., $R=$ $\frac{|m|}{\sqrt{1+k^{2}}}$, i.e., $m^{2}=\frac{8}{3}\left(1+k^{2}\right)$, substituting into (4), we get $b^{2}=4$.

When the slope of line $l$ does not exist, the equation of $l$ can be set as $x=\sqrt{\frac{8}{3}}$, substituting into the ellipse equation, we get $y= \pm \sqrt{\frac{2}{3}} b$, by (3), we know $\frac{8}{3}-\frac{2}{3} b^{2}=0, b^{2}=4$.
In summary, regardless of whether the slope of line $l$ exists, we have $b^{2}=4$.
(2) Draw $O D \perp A B$ from the origin $O$, with the foot of the perpendicular being $D$, then $D$ is the point of tangency, then $O D=$ $R=\frac{2 \sqrt{6}}{3}$, let $\angle O A B=\theta$, then $\theta$ is an acute angle, and $|A D|=\frac{R}{\tan \theta}=\frac{2 \sqrt{6}}{3 \tan \theta}$, $|B D|=\frac{2 \sqrt{6}}{3} \tan \theta$, so
$$
|A B|=|A D|+|B D|=\frac{2 \sqrt{6}}{3}\left(\tan \theta+\frac{1}{\tan \theta}\right) .
$$

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, so when $x=$ 1, $|A B|_{\text {min }}=\frac{4 \sqrt{6}}{3}$;
When $x=\sqrt{2}$ (or $\frac{\sqrt{2}}{2}$), $|A B|_{\max }=2 \sqrt{3}$, i.e., $\frac{4 \sqrt{6}}{3} \leqslant|A B| \leqslant 2 \sqrt{3}$.