Axis), $F$ is the focus, and $|A F|+|B F|=8$. The perpendicular bisector of line segment $A B$ always passes through the fixed point $Q(6,0)$ on the $x$-axis.
(1) Find the equation of the parabola $C$;
(2) Find the maximum value of the area of $\triangle A Q B$.

保留了源文本的换行和格式。
(1) Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ and $x_{1} \neq x_{2}$.
By the property of the parabola: $8-|A F|+|B F|=x_{1}+\frac{P}{2}+x_{2}+\frac{P}{2}=x_{1}+x_{2}+P$, and since $|A Q|=|B Q|$, we have $\left(x_{1}-6\right)^{2}+y_{1}^{2}=\left(x_{2}-6\right)^{2}+y_{2}^{2}$, so, $\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}+2 P-12\right)=0$, and since $x_{1} \neq x_{2}$, then $P+8-12=0$, i.e., $P=4$. Therefore, the equation of the parabola is: $y^{2}=8 x$.
(2) From (1), we know $x_{1}+x_{2}=4$, we can set the coordinates of the midpoint $E$ of segment $A B$ as $\left(2, y_{0}\right), y_{0} \neq 0$, then $\left\{\begin{array}{l}y-y_{0}=\frac{4}{y_{0}}(x-2), \\ y^{2}=8 x\end{array} \Rightarrow y^{2}-2 y_{0} y+2 y_{0}^{2}-16=0\right.$, so, $\Delta=4 y_{0}^{2}-4\left(2 y_{0}^{2}-16\right)=4\left(16-y_{0}^{2}\right)>0$, solving gives $y_{0} \in(-4,0) \cup(0,4)$.
Therefore,
$$
\begin{aligned}
S_{\triangle A Q B} & =\frac{1}{2}|A B| \cdot|E Q|=\frac{1}{2} \sqrt{1+\left(\frac{y_{0}}{4}\right)^{2}} \sqrt{4\left(16-y_{0}^{2}\right)} \sqrt{y_{0}^{2}+16} \\
& =\frac{1}{4 \sqrt{2}} \sqrt{\left(32-2 y_{0}^{2}\right)\left(16+y_{0}^{2}\right)\left(16+y_{0}^{2}\right)} \\
& \leqslant \frac{1}{4 \sqrt{2}} \sqrt{\left(\frac{64}{3}\right)^{3}}=\frac{64}{9} \sqrt{6},
\end{aligned}
$$

When and only when $32-2 y_{0}^{2}=16+y_{0}^{2}$, i.e., $y_{0}= \pm \frac{4}{3} \sqrt{3}$, the maximum value is reached. The maximum value of $S_{\triangle A Q B}$ is $\frac{64}{9} \sqrt{6}$.