11. (This question is worth 20 points) Let the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively, and the right vertex be $A$. Let $P$ be any point on the ellipse $C_{1}$, and the maximum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is in the range $\left[c^{2}, 3 c^{2}\right]$, where $c=\sqrt{a^{2}-b^{2}}$.
(1) Find the range of the eccentricity $e$ of the ellipse $C_{1}$:
(2) Let the hyperbola $C_{2}$ have the foci of the ellipse $C_{1}$ as its vertices and the vertices of the ellipse $C_{1}$ as its foci. Let $B$ be any point on the hyperbola $C_{2}$ in the first quadrant. When $e$ takes its minimum value, does there exist a constant $\lambda(\lambda>0)$ such that $\angle B A F_{1}=\lambda \angle B F_{1} A$ always holds? If it exists, find the value of $\lambda$; if not, explain the reason.
Solution: (1) Let $P(x, y)$, then $\overrightarrow{P F_{1}}=(-c-x,-y), \overrightarrow{P F_{2}}=(c-x,-y)$.
$\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x^{2}+y^{2}-c^{2}$. Also, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, so $y^{2}=b^{2}-\frac{b^{2} x^{2}}{a^{2}}, 0 \leq x^{2} \leq a^{2}$.
Therefore, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=\left(1-\frac{b^{2}}{a^{2}}\right) x^{2}+b^{2}-c^{2}=\frac{c^{2}}{a^{2}} x^{2}+b^{2}-c^{2}$
Thus, when $x^{2}=a^{2}$, $\left.\overrightarrow{P F}_{1} \cdot \overrightarrow{P F_{2}}\right|_{\text {max }}=b^{2}$, so the range of $b^{2}$ is $\left[c^{2}, 3 c^{2}\right]$.
From this, it is easy to see that $\frac{1}{2} \leq e \leq \frac{\sqrt{2}}{2}$.
(2) When $e=\frac{1}{2}$, $a=2 c, b=\sqrt{3} c$, then $C_{2}: \frac{x^{2}}{c^{2}}-\frac{y^{2}}{3 c^{2}}=1, A(2 c, 0)$, let $B\left(x_{0}, y_{0}\right),\left(x_{0}>0, y_{0}>0\right)$, then $\frac{x_{0}{ }^{2}}{c^{2}}-\frac{y_{0}{ }^{2}}{3 c^{2}}=1$.
Guess $\lambda=2$, such that $\angle B A F_{1}=\lambda \angle B F_{1} A$ always holds.
When $A B \perp x$-axis, it is obviously true.
When $x_{0} \neq 2 c$, $\tan \angle B A F_{1}=\frac{-y_{0}}{x_{0}-a}=\frac{-y_{0}}{x_{0}-2 c}, \tan \angle B F_{1} A=\frac{y_{0}}{x_{0}+c}$.
Therefore, $\tan 2 \angle B F_{1} A=\frac{2 \tan \angle B F_{1} A}{1-\tan ^{2} \angle B F_{1} A}=\frac{\frac{2 y_{0}}{x_{0}+c}}{1-\left(\frac{y_{0}}{x_{0}+c}\right)^{2}}$.
Since $y_{0}{ }^{2}=3 c^{2}\left(\frac{x_{0}{ }^{2}}{c^{2}}-1\right)=3\left(x_{0}^{2}-c^{2}\right)$,
$\tan 2 \angle B F_{1} A=\frac{2 y_{0}\left(x_{0}+c\right)}{\left(x_{0}+c\right)^{2}-y_{0}{ }^{2}}=\frac{2 y_{0}\left(x_{0}+c\right)}{\left(x_{0}+c\right)^{2}-3\left(x_{0}{ }^{2}-c^{2}\right)}=\frac{y_{0}}{2 c-x_{0}}=\tan \angle B A F_{1}$ and $2 \angle B F_{1} A$ and $\angle B A F_{1}$ are both in $(0, \pi)$,
Therefore, $2 \angle B F_{1} A=\angle B A F_{1}$,
So when $\lambda=2$, $\angle B A F_{1}=\lambda \angle B F_{1} A$ always holds.
Proof completed!