11. Given the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>1)$ with an eccentricity of $\frac{2}{3}$, a circle $K$ is drawn with the left vertex $K$ as the center and $\sqrt{6}$ as the radius. The circle $K$ intersects the ellipse $\Gamma$ at points $A$ and $B$, and the midpoint of segment $A B$ has an x-coordinate of $-\frac{3}{2}$.
(1) Find the equation of the ellipse $\Gamma$;
(2) The circle $K$ intersects the line $x=-\frac{9}{2}$ at points $C$ and $D$. Point $P$ lies on segment $C D$. Two different tangents $P E$ and $P F$ (with $E$ and $F$ as the points of tangency) are drawn from point $P$ to the ellipse $\Gamma$. Find the range of the area of $\triangle P E F$.
(1) As shown in the figure, $e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{4}{9} \Rightarrow \frac{b^{2}}{a^{2}}=\frac{5}{9}$.

Let $K(-a, 0), A\left(-\frac{3}{2}, y_{0}\right)$,
then $|A K|^{2}=\left(a-\frac{3}{2}\right)^{2}+y_{0}^{2}$
$=a^{2}-3 a+\frac{9}{4}+b^{2}\left(1-\frac{x_{0}^{2}}{a^{2}}\right)$
$$
=a^{2}+b^{2}-3 a+\frac{9}{4}-\frac{5}{9} \cdot \frac{9}{4}=6 \Rightarrow \frac{14}{9} a^{2}-3 a-5=0 \Rightarrow a=3, b=\sqrt{5} \text {. }
$$

Therefore, the equation of the ellipse $\Gamma$ is $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$.
(2) Let $P\left(-\frac{9}{2}, t\right), E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, the tangent line equation of the ellipse $\Gamma$ at point $E$ is $\frac{x_{1} x}{9}+\frac{y_{1} y}{5}=1$, similarly, the tangent line equation of the ellipse $\Gamma$ at point $F$ is $\frac{x_{2} x}{9}+\frac{y_{2} y}{5}=1$. Substituting point $P$ into these equations, we get the equation of line $E F$ as $-\frac{x}{2}+\frac{t y}{5}=1 \Rightarrow x=\frac{2 t}{5} y-2$, which means $E F$ passes through the fixed point $G(-2,0)$, and $k_{P G}=\frac{t}{-\frac{9}{2}+2}=-\frac{2 t}{5} \Rightarrow P G \perp E F$.

It is easy to know that $t \in\left[-\frac{\sqrt{15}}{2}, \frac{\sqrt{15}}{2}\right]$, solving the system $\left\{\begin{array}{l}x=\frac{2 t}{5} y-2, \\ 5 x^{2}+9 y^{2}=45\end{array} \Rightarrow\left(4 t^{2}+45\right) y^{2}-40 t y-125=0\right.$,
thus $|E F|=\sqrt{1+\left(\frac{2 t}{5}\right)^{2}} \cdot \frac{\sqrt{40^{2} t^{2}+4 \cdot 125\left(4 t^{2}+45\right)}}{4 t^{2}+45},|P G|=\frac{5}{2} \sqrt{1+\left(\frac{2 t}{5}\right)^{2}}$.
Therefore, $S_{\triangle P E F}=\frac{5}{4}\left[1+\left(\frac{2 t}{5}\right)^{2}\right] \cdot \frac{\sqrt{40^{2} t^{2}+4 \cdot 125\left(4 t^{2}+45\right)}}{4 t^{2}+45}$
$=\left(4 t^{2}+25\right) \cdot \frac{\sqrt{16 t^{2}+5\left(4 t^{2}+45\right)}}{2\left(4 t^{2}+45\right)}=\frac{3\left(4 t^{2}+25\right) \sqrt{4 t^{2}+25}}{2\left(4 t^{2}+45\right)} \xlongequal{m=\sqrt{4 t^{2}+25} \in[5,2 \sqrt{10}]} \frac{3 m^{3}}{2\left(m^{2}+20\right)}$
$=\frac{3}{2\left(\frac{1}{m}+\frac{20}{m^{3}}\right)}$ is monotonically increasing with respect to $m$, so the range of the area of $\triangle P E F$ is $\left[\frac{25}{6}, 2 \sqrt{10}\right]$.