2. (15 points) Given the hyperbola $S: x^{2}-4 y^{2}=8$, and a moving chord $A B$ on the right branch of $S$, with the midpoint of $A B$ being $C$.
(1) If the distance from $C$ to one of the asymptotes $l_{1}$ of the hyperbola that passes through the first and third quadrants is 1, and the other asymptote $l_{2}$ intersects $A B$ at point $D$, with $O$ being the origin, find the length of $O D$;
(2) If $A B=4$, and $A B$ intersects $l_{1}$ and $l_{2}$ at points $P$ and $D$, respectively, find the range of $S_{\triangle P O D}$.
2. (1) When $AB$ is not a perpendicular chord, let $l_{AB}: y=kx+b$. Combining with $S$ we get
$\left(1-4k^{2}\right)x^{2}-8kbx-4b^{2}-8=0$.
From $AB$ intersecting the hyperbola at two points, we know
$k \neq \pm \frac{1}{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we get $x_{1} \star x_{2}=\frac{8kb}{1-4k^{2}}$.
Let the midpoint of $AB$ be $C\left(x_{0}, y_{0}\right)$. Then
$x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{4kb}{1-4k^{2}}$,
$y_{0}=\frac{y_{1}+y_{2}}{2}=\frac{kx_{1}+b+kx_{2}+b}{2}=\frac{b}{1-4k^{2}}$.
Also, the distance from point $C$ to $l_{1}: x-2y=0$ is 1, so $\frac{\left|x_{0}-2y_{0}\right|}{\sqrt{5}}=1$, which means $\left|\frac{b}{1+2k}\right|=\frac{\sqrt{5}}{2}$.
It is easy to find that the intersection point of $l_{2}: x+2y=0$ and $AB$ is $D\left(\frac{-2b}{2k+1}, \frac{b}{2k+1}\right)$.
Thus, $|OD|=\sqrt{\left(\frac{-2b}{2k+1}\right)^{2}+\left(\frac{b}{2k+1}\right)^{2}}$ $=\sqrt{5}\left|\frac{b}{2k+1}\right|=\frac{5}{2}$.
When $AB \perp x$-axis, it is easy to get $|OD|=\frac{5}{2}$.
Therefore, $|OD|=\frac{5}{2}$.
(2) From (1), when $AB$ is not a perpendicular chord,
$x_{1}+x_{2}=\frac{8kb}{1-4k^{2}}, x_{1}x_{2}=\frac{-4\left(b^{2}+2\right)}{1-4k^{2}}$.
Thus, $|AB|^{2}=\left(1+k^{2}\right)\left(x_{1}-x_{2}\right)^{2}$
$=\left(1+k^{2}\right)\left[\left(x_{1}+x_{2}\right)^{2}-4x_{1}x_{2}\right]$.
Since $AB=4$, rearranging gives $b^{2}=\frac{\left(6k^{2}+1\right)\left(4k^{2}-1\right)}{1+k^{2}}$.
Thus, it is easy to see that $k^{2}>\frac{1}{4}$.
Also, $AB$ intersects $l_{1}$ and $l_{2}$ at points $P$ and $D$, and it is easy to calculate $P\left(\frac{2b}{1-2k}, \frac{b}{1-2k}\right)$ and $D\left(\frac{-2b}{2k+1}, \frac{b}{2k+1}\right)$.
Thus, $|PD|=\frac{4|b|\sqrt{1+k^{2}}}{4k^{2}-1}$.
The distance from point $O$ to $PD$ is $\frac{|b|}{\sqrt{1+k^{2}}}$.
Therefore, $S_{\triangle POD}=\frac{1}{2} \cdot \frac{|b|}{\sqrt{1+k^{2}}} \cdot \frac{4|b|\sqrt{1+k^{2}}}{4k^{2}-1}$ $=\frac{2b^{2}}{4k^{2}-1}=\frac{2\left(6k^{2}+1\right)}{1+k^{2}}$.
Since $k^{2}>\frac{1}{4}$, we know $S_{\triangle POD}>4, S_{\triangle POD}<12$.
When $AB \perp x$-axis, $S_{\triangle POD}=12$.
Thus, the range of $S_{\triangle POD}$ is $(4,12]$.