Given $F$ and $l$ as the corresponding focus and directrix, draw a line through $F$ parallel to $y=x$ intersecting the ellipse $C$ at points $A$ and $B$. It is known that the center $P$ of $C$ is inside the circle with $A B$ as its diameter. Find the range of the eccentricity $e$ of the ellipse $C$.
From $3 x^{2}-y^{2}-24 x+36=0$, we get $\frac{(x-4)^{2}}{4}-\frac{y^{2}}{12}=1$. Therefore, the center of the hyperbola $\frac{(x-4)^{2}}{4}-\frac{y^{2}}{12}-1$ is $(4,0)$, the right focus is $F(8,0)$, and $l: x=4+\frac{4}{4}=5$.

Let $M(x, y)$ be any point on the ellipse $C$, $2 a$ be the length of the major axis of the ellipse, and $2 c$ be the focal distance of the ellipse. Let $A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right)$, then $\frac{\sqrt{(x-8)^{2}+y^{2}}}{|x-5|}=e$, (1) and the equation of the line $A B$ is $y=$ $x-8$, (2) From (1) and (2), we get $(2- \left.e^{2}\right) x^{2}-\left(32-10 e^{2}\right) x+128-5 e^{2}=0$. By the problem, $x_{A}$ and $x_{B}$ are the two roots of this equation, so
$$
\begin{array}{c}
x_{A}+x_{B}=\frac{32-10 e^{2}}{2-e^{2}}=16+\frac{6 e^{2}}{2-e^{2}}, x_{A} \cdot x_{B}=\frac{128-5 e^{2}}{2-e^{2}}, \\
y_{A}+y_{B}=x_{A}-8+x_{B}-8=\frac{32-10 e^{2}}{2-e^{2}}-16=\frac{6 e^{2}}{2-e^{2}},
\end{array}
$$

The coordinates of the center are $\left(8+\frac{3 e^{2}}{2-e^{2}}, \frac{3 e^{2}}{2-e^{2}}\right)$,
$$
\begin{aligned}
|A B| & =\sqrt{2}\left|x_{A}-x_{B}\right|=\sqrt{2} \cdot \sqrt{\left(x_{A}+x_{B}\right)^{2}-4 x_{A} x_{B}} \\
& =\sqrt{2} \cdot \sqrt{\left(\frac{32-10 e^{2}}{2-e^{2}}\right)^{2}-4 \frac{128-25 e^{2}}{2-e^{2}}}=\frac{12 e}{2-e^{2}} .
\end{aligned}
$$

In the ellipse $C$, from $\frac{a-c}{3-(a-c)}=e$, we get $a-c=\frac{3 e}{e+1}$, and $\frac{c}{a}=e$, so $c=\frac{3 e^{2}}{1-e^{2}}$, hence the center coordinates of the ellipse are $P\left(8+\frac{3 e^{2}}{1-e^{2}}, 0\right)$, and $P$ is inside the circle with $A B$ as its diameter, so
$$
\left(\frac{3 e^{2}}{1-e^{2}}-\frac{3 e^{2}}{2-e^{2}}\right)^{2}+\left(\frac{3 e^{2}}{2-e^{2}}\right)^{2}0, e^{2}-2<-\sqrt{2}$, i.e., $e^{2}<2-\sqrt{2}$, thus $0<e<\sqrt{2-\sqrt{2}}$.