Intersecting the hyperbola $y^{2}-x^{2}=1$ at two points each, respectively $A_{1} 、 B_{1}$ and $A_{2} 、 B_{2}$.
(1) Find the range of the slope $k_{1}$ of $l_{1}$;
(2) If $\left|A_{1} B_{1}\right|=\sqrt{5}\left|A_{2} B_{2}\right|$, find the equations of $l_{1} 、 l_{2}$.
(1) According to the problem, let the slopes of $l_{1}$ and $l_{2}$ both exist. Since $l_{1}$ passes through point $P(-\sqrt{2}, 0)$ and intersects the hyperbola at two points, the system of equations 
$$
\left\{\begin{array}{l}y=k_{1}(x+\sqrt{2}) \\ y^{2}-x^{2}=1,\end{array} \quad\left(k_{1} \neq 0\right)\right.
$$
(1) has two distinct solutions. Eliminating $y$ and rearranging gives
$$
\left(k_{1}^{2}-1\right) x^{2}+2 \sqrt{2} k_{1}^{2} x+2 k_{1}^{2}-1=0,
$$

If $k_{1}^{2}-1=0$, then the system of equations (1) has only one solution, which does not meet the problem's requirements. Therefore, $k_{1}^{2}-1 \neq 0$ and the discriminant of equation (2) is $\Delta_{1}=4\left(3 k_{1}^{2}-1\right)>0$.
Let the slope of $l_{2}$ be $k_{2}$. It is known that the system of equations 
$$
\left\{\begin{array}{l}y=k_{2}(x+\sqrt{2}) \quad\left(k_{2} \neq 0\right), \\ y^{2}-x^{2}=1,\end{array}\right.
$$
(3) has two distinct solutions. Eliminating $y$ and rearranging gives
$$
\left(k_{2}^{2}-1\right) x^{2}+2 \sqrt{2} k_{2}^{2} x+2 k_{2}^{2}-1=0,
$$

Similarly, $k_{2}^{2}-1 \neq 0$, and the discriminant of equation (4) is $\Delta_{2}=4\left(3 k_{2}^{2}-1\right)>0$.
Since $l_{1}$ and $l_{2}$ are perpendicular, $k_{1} k_{2}=-1$. Therefore, according to the problem, we have 
$$
\left\{\begin{array}{l}3 k_{1}^{2}-1>0, \\ 3 k_{2}^{2}-1>0, \\ k_{1} \cdot k_{2}=-1, \\ \left|k_{1}\right| \neq 1 .\end{array}\right.
$$
Solving this, we get 
$$
\left\{\begin{array}{l}\frac{\sqrt{3}}{3}<\left|k_{1}\right|<\sqrt{3}, \\ \left|k_{1}\right| \neq 1 .\end{array}\right.
$$
Thus, $k_{1} \in(-\sqrt{3},-1) \cup\left(-1,-\frac{\sqrt{3}}{3}\right) \cup\left(\frac{\sqrt{3}}{3}, 1\right) \cup(1, \sqrt{3})$.
(2) Let the coordinates of $A_{1}$ and $B_{1}$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. From equation (2), we know
$$
x_{1}+x_{2}=\frac{-2 \sqrt{2} k_{1}^{2}}{k_{1}^{2}-1}, x_{1} \cdot x_{2}=\frac{2 k_{1}^{2}-1}{k_{1}^{2}-1},
$$

Therefore, 
$$
\left|A_{1} B_{1}\right|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=\left(1+k_{1}^{2}\right)\left(x_{1}-x_{2}\right)^{2}=\frac{4\left(1+k_{1}^{2}\right)\left(3 k_{1}^{2}-1\right)}{\left(k_{1}^{2}-1\right)^{2}}.
$$

Similarly, from equation (4), we can find 
$$
\left|A_{2} B_{2}\right|^{2}=\frac{4\left(1+k_{2}^{2}\right)\left(3 k_{2}^{2}-1\right)}{\left(k_{2}^{2}-1\right)^{2}}.
$$
(6) From $\left|A_{1} B_{1}\right|=\sqrt{5}\left|A_{2} B_{2}\right|$, we get $\left|A_{1} B_{1}\right|^{2}=5\left|A_{2} B_{2}\right|^{2}$. Substituting (5) and (6) into the above equation, we get
$$
\text { - } \frac{4\left(1+k_{1}^{2}\right)\left(3 k_{1}^{2}-1\right)}{\left(k_{1}^{2}-1\right)^{2}}=5 \times \frac{4\left(1+k_{2}^{2}\right)\left(3 k_{2}^{2}-1\right)}{\left(k_{2}^{2}-1\right)^{2}}.
$$

Combining $k_{1} k_{2}=-1$, we solve to get $k_{1}= \pm \sqrt{2}$.
Taking $k_{1}=\sqrt{2}$, we get $l_{1}: y=\sqrt{2}(x+\sqrt{2}), l_{2}: y=-\frac{\sqrt{2}}{2}(x+\sqrt{2})$; taking $k_{1}=-\sqrt{2}$, we get $l_{1}:$ $y=-\sqrt{2}(x+\sqrt{2}), l_{2}: y=\frac{\sqrt{2}}{2}(x+\sqrt{2})$.