1's common left and right vertices, $P$ and $Q$ are moving points on the hyperbola and ellipse, respectively, different from $A$ and $B$, and satisfy: $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\widehat{A Q}+\overrightarrow{B Q})(\lambda \in \mathbf{R}$ and $|\lambda|>1)$. Let the slopes of the lines $A P$, $B P$, $A Q$, and $B Q$ be $k_{1}$, $k_{2}$, $k_{3}$, and $k_{4}$, respectively.
(1) Prove: $k_{1}+k_{2}+k_{3}+k_{4}=0$;
(2) Let $F_{1}$ and $F_{2}$ be the right foci of the ellipse and hyperbola, respectively. If $P F_{2} \parallel Q F_{1}$, find the value of $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}$.
(1) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and we know $A(-a, 0), B(a, 0)$, then we have
$$
\begin{array}{c}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \frac{x_{1}}{y_{1}}, \\
k_{3}+k_{4}=\frac{y_{2}}{x_{2}+a}+\frac{y_{2}}{x_{2}-a}=\frac{2 x_{2} y_{2}}{x_{2}^{2}-a^{2}}=-\frac{2 b^{2}}{a^{2}} \frac{x_{2}}{y_{2}} .
\end{array}
$$

Let $O$ be the origin, then $2 \overrightarrow{O P}=\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})=2 \lambda \cdot \overrightarrow{O Q}$, so $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$, hence $O, P, Q$ are collinear, thus $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$, from (1) and (2) we can get $k_{1}+k_{2}+k_{3}+k_{4}=0$.
(2) Since point $Q$ is on the ellipse, we have $\frac{x_{2}^{2}}{a^{2}}+\frac{y_{2}^{2}}{b^{2}}=1$, and from (1) we know $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$, so $\left(x_{1}, y_{1}\right)=$ $\lambda\left(x_{2}, y_{2}\right)$, thus $x_{2}=\frac{1}{\lambda} x_{1}, y_{2}=\frac{1}{\lambda} y_{1}$, hence $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=\lambda^{2}$,
(3) Since point $P$ is on the hyperbola, we have $\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1$, (4) From (3) and (4) we can get $x_{1}^{2}=\frac{\lambda^{2}+1}{2} a^{2}, y_{1}^{2}=\frac{\lambda^{2}-1}{2} b^{2}$. Since $P F_{2} \parallel Q F_{1}$, we have $\frac{\left|O F_{2}\right|}{\left|O F_{1}\right|}=\frac{|O P|}{|O Q|}=\lambda$, thus $\lambda^{2}=\frac{\left|O F_{2}\right|^{2}}{\left|O F_{1}\right|^{2}}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$, hence
$$
\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{\lambda^{2}+1}{\lambda^{2}-1} \cdot \frac{a^{2}}{b^{2}}=\frac{a^{4}}{b^{4}} .
$$

From (1) we get $\left(k_{1}+k_{2}\right)^{2}-\frac{4 b^{4}}{a^{4}} \frac{x_{1}^{2}}{y_{1}^{2}}-4$, similarly $\left(k_{3}+k_{4}\right)^{2}=4$. On the other hand, $k_{1} k_{2}=$ $\frac{y_{1}}{x_{1}+a} \cdot \frac{y_{1}}{x_{1}-a}=\frac{y_{1}^{2}}{x_{1}^{2}-a^{2}}=\frac{b^{2}}{a^{2}}, k_{3} k_{4}=\frac{y_{2}}{x_{2}+a} \cdot \frac{y_{2}}{x_{2}-a}=\frac{y_{2}^{2}}{x_{2}^{2}-a^{2}}=-\frac{b^{2}}{a^{2}}$, thus
$$
k_{1}^{2}+k_{2}^{2}+k_{3}^{2}+k_{4}^{2}=\left(k_{1}+k_{2}\right)^{2}+\left(k_{3}+k_{4}\right)^{2}-2\left(k_{1} k_{2}+k_{3} k_{4}\right)=8 .
$$