11. (20 points) In the Cartesian coordinate system $x O y$, a line $l$ is drawn through the point $M(m, 0) (m>0)$, intersecting the parabola $y^{2}=2 p x (p>0)$ at points $A$ and $B$. A line $l^{\prime}$ perpendicular to the $x$-axis is drawn through the point $M^{\prime}$, which is the symmetric point of $M$ with respect to the origin.
(1) Let the intersection points of the lines $A O$ and $B O$ with $l^{\prime}$ be $C$ and $D$, respectively. Discuss the positional relationship between the lines $A D$ and $B C$ and the $x$-axis;
(2) Let the focus of the parabola $y^{2}=2 p x$ be $F$. If $m=p$ and $\angle A M^{\prime} O=30^{\circ}$, find the size of $\angle A F B$.
11. (1) Lines $A D$ and $B C$ are both parallel to the $x$-axis. Let the slope of line $l$ be $k$, and assume $k>0$. Then the equation of line $l$ is $y=k(x-m)$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$,
Clearly, $y_{1}$ and $y_{2}$ have opposite signs.
First, we prove: $\angle A M^{\prime} O=\angle B M^{\prime} O$.
Equation (1) $\Leftrightarrow \tan \angle A M^{\prime} O=\tan \angle B M^{\prime} O$
$$
\begin{array}{c}
\Leftrightarrow \frac{y_{1} \cdots}{x_{1}+m}=-\frac{y_{2}}{x_{2}+m} \\
\Leftrightarrow x_{1} y_{2}+x_{2} y_{1}+m\left(y_{1}+y_{2}\right)=0 . \\
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
y=k(x-m)
\end{array} \text { eliminating } y\right. \text { and rearranging, we get } \\
k^{2} x^{2}-2\left(k^{2} m+p\right) x+k^{2} m^{2}=0 .
\end{array}
$$

From $\left\{\begin{array}{l}y^{2}=2 p x, \\ y=k(x-m)\end{array}\right.$ eliminating $y$ and rearranging, we get

By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=2\left(m+\frac{p}{k^{2}}\right), \\
x_{1} x_{2}=m^{2} .
\end{array}
$$

Then $y_{1}+y_{2}=k\left(x_{1}+x_{2}-2 m\right)=\frac{2 p}{k}$,
$y_{1} y_{2}=-2 p \sqrt{x_{1} x_{2}}=-2 p m$.
Thus, $x_{1} y_{2}+x_{2} y_{1}+m\left(y_{1}+y_{2}\right)$
$$
=2 k x_{1} x_{2}-k m\left(x_{1}+x_{2}\right)+m\left(y_{1}+y_{2}\right)=0 \text {. }
$$

From equation (2), we get
$$
\begin{array}{l}
\frac{m}{x_{2}}=\frac{x_{1}}{m} \Rightarrow \frac{x_{1}-m}{m-x_{2}}=\frac{x_{1}}{m} \\
\Rightarrow \frac{A M}{B M}=\frac{x_{1}-m}{m-x_{2}}=\frac{x_{1}}{m}=\frac{O A}{O C} \Rightarrow O M / / B C .
\end{array}
$$

Similarly, $O M / / A D$.
(2) From (1), we know
$\angle B M^{\prime} O=\angle A M^{\prime} O=30^{\circ}$,
which means $\frac{y_{1}}{x_{1}+p}=\frac{\sqrt{3}}{3}, \frac{y_{2}}{x_{2}+p}=-\frac{\sqrt{3}}{3}$.
Then $y_{1} y_{2}=-\frac{1}{3}\left[x_{1} x_{2}+p\left(x_{1}+x_{2}\right)+p^{2}\right]$
$$
=-\frac{2 p^{2}}{3}\left(2+\frac{1}{k^{2}}\right) \text {. }
$$

Combining $y_{1} y_{2}=-2 p^{2}$, we solve to get $k=1$.
Thus, $\cos \angle A F B$
$$
\begin{array}{l}
=\frac{\left(x_{1}+\frac{p}{2}\right)^{2}+\left(x_{2}+\frac{p}{2}\right)^{2}-2\left(x_{1}-x_{2}\right)^{2}}{2\left(x_{1}+\frac{p}{2}\right)\left(x_{2}+\frac{p}{2}\right)} \\
=\frac{p\left(x_{1}+x_{2}\right)+\frac{p}{2}-\left(x_{1}+x_{2}\right)^{2}+6 x_{1} x_{2}}{2 x_{1} x_{2}+p\left(x_{1}+x_{2}\right)+\frac{p^{2}}{2}} \\
=\frac{4+\frac{1}{2}-16+6}{2+4+\frac{1}{2}}=-\frac{11}{13} .
\end{array}
$$

Therefore, $\angle A F B=\pi-\arccos \frac{11}{13}$.