II. (25 points) As shown in Figure 4, given the parabola
$$
y=a x^{2}+b x+8\left(a>\frac{1}{2}\right)
$$

passes through point $D(5,3)$, intersects the $x$-axis at points $B\left(x_{1}, 0\right)$ and $C\left(x_{2}, 0\right)$, and $S_{\triangle B C D}=3$. A line $l \perp C D$ is drawn through point $D$ and intersects the $y$-axis at point $A$ and the $x$-axis at point $P$.
(1) Find the equation of the parabola and the value of $P D \cdot P A$;
(2) Let $B D$ intersect $A C$ at point $Q$, find the value of $\frac{A Q}{Q C}$.
(1) From $S_{\triangle B C D}=3$, we get
$$
\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-32 a}}{|a|}=2 \text {. }
$$

Also, from $25 a+5 b+8=3$, we get $a=1, b=-6$.
Thus, the equation of the parabola is $y=x^{2}-6 x+8$.
It is easy to find that $C(4,0)$.
Let $P(t, 0)$.
As shown in Figure 9, draw $D E \perp x$-axis at point $E$. Then $E(5,0)$.
Since $C D \perp A P \Rightarrow D E^{2}=C E \cdot P E \Rightarrow t=14$.
Also, $l \perp C D$
$\Rightarrow O, C, D, A$ are concyclic
$$
\Rightarrow P D \cdot P A=P C \cdot P O=140 \text {. }
$$
(2) Note that,
$$
\begin{array}{l}
E D / / O A \Rightarrow \frac{E D}{O A}=\frac{E P}{O P} \Rightarrow O A=\frac{14}{3} \\
\Rightarrow l_{A C}: y=-\frac{7}{6} x+\frac{14}{3} .
\end{array}
$$

And $l_{B D}: y=x-2$.
Solving the two equations simultaneously, we get $x=\frac{40}{13}, y=\frac{14}{13}$.
$$
\text { Hence } \frac{A Q}{Q C}=\frac{\frac{40}{13}}{4-\frac{40}{13}}=\frac{10}{3} \text {. }
$$