The right vertex is $A, P$ is 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) The range of the eccentricity $e$ of the ellipse $C_{1}$;
(2) Suppose the hyperbola $C_{2}$ has the foci of the ellipse $C_{1}$ as its vertices and the vertices of the ellipse $C_{1}$ as its foci. $B$ is 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.
(1) Let $P(x, y)$, and $F_{1}(-c, 0), F_{2}(c, 0)$, so
$$
\overrightarrow{P F_{1}}=(-c-x,-y), \overrightarrow{P F_{2}}=(c-x,-y),
$$

Thus, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x^{2}+y^{2}-c^{2}$. Also, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, which gives $y^{2}=b^{2}-\frac{b^{2} x^{2}}{a^{2}}, 0 \leqslant x^{2} \leqslant a^{2}$. Therefore,
$$
\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=\left(1-\frac{b^{2}}{a^{2}}\right) x^{2}+b^{2}-c^{2}=\frac{c^{2}}{a^{2}} x^{2}+b^{2}-c^{2} .
$$

Hence, when $x^{2}=a^{2}$, $\left(\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}\right)_{\max }=b^{2}, c^{2} \leqslant b^{2} \leqslant 3 c^{2}, c^{2} \leqslant a^{2}-c^{2} \leqslant 3 c^{2}$. Therefore, $\frac{1}{4} \leqslant \frac{c^{2}}{a^{2}} \leqslant \frac{1}{2}$, which means $\frac{1}{4} \leqslant e^{2} \leqslant \frac{1}{2}$. Thus, $\frac{1}{2} \leqslant e \leqslant \frac{\sqrt{2}}{2}$.
(2) When $e=\frac{1}{2}$, $a=2 c, b=\sqrt{3} c$. Therefore, $C_{2}: \frac{x^{2}}{c^{2}}-\frac{y^{2}}{3 c^{2}}=1, A(2 c, 0)$.
Let $B\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$, then $\frac{x_{0}^{2}}{c^{2}}-\frac{y_{0}^{2}}{3 c^{2}}=1$.
When $A B \perp x$-axis, $x_{0}=2 c, y_{0}=3 c$, then $\tan \angle B F_{1} A=\frac{3 c}{3 c}=1$. Hence, $\angle B F_{1} A=\frac{\pi}{4}$. Thus, $\angle B A F_{1}=\frac{\pi}{2}=2 \angle B F_{1} A$, leading to the conjecture that $\lambda=2$, such that $\angle B A F_{1}=\lambda \angle B F_{1} A$ always holds.
When $x_{0} \neq 2 c$, $\tan \angle B A F_{1}=\frac{-y_{0}}{x_{0}-a}=\frac{-y_{0}}{x_{0}-2 c}, \tan \angle B F_{1} \Lambda=\frac{y_{0}}{x_{0}+c}$. Therefore,
$$
\tan 2 \angle B F_{1} A=\frac{2 \tan \angle B F_{1} A}{1-\tan ^{2} \angle B F_{1} A}=\frac{\frac{2 y_{0}}{x_{0}+c}}{1-\left(\frac{y_{0}}{x_{0}+c}\right)^{2}} \text {. }
$$

Also, $y_{0}^{2}=3 c^{2}\left(\frac{x_{0}^{2}}{c^{2}}-1\right)=3\left(x_{0}^{2}-c^{2}\right)$, so
$$
\tan 2 \angle B F_{1} A=\frac{2 y_{0}\left(x_{0}+c\right)}{\left(x_{0}+c\right)^{2}-3\left(x_{0}^{2}-c^{2}\right)}=\frac{-y_{0}}{x_{0}-2 c}=\tan \angle B A F_{1} .
$$

Since $2 \angle B F_{1} A$ and $\angle B A F_{1}$ are both in $\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)$, it follows that $2 \angle B F_{1} A=$ $\angle B A F_{1}$, hence there exists $\lambda=2$, such that $\angle B A F_{1}=\lambda \angle B F_{1} A$ always holds.