- Four, (20 points) Given that $P$ and $Q$ are two moving points on the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $O$ is the origin, the slope of line $O P$ is $k$, the product of the slopes of lines $O P$ and $O Q$ is $m$, and $p=|O P|^{2}+|O Q|^{2}$ is a constant independent of $k$.
(1) Find the values of $m$ and $p$;
(2) If the hyperbola $\Gamma$ has its foci on the $x$-axis and asymptotes given by $y= \pm m x$, and the eccentricities of the ellipse $C$ and the hyperbola $\Gamma$ are $e_{1}$ and $e_{2}$ respectively, find the range of $e_{2}-e_{1}$.
(1) The equation of the line $O P$ is $y=k x$, and when combined with the equation of the ellipse $C$, we get
$$
x^{2}=\frac{a^{2} b^{2}}{b^{2}+a^{2} k^{2}} \text {. }
$$

Then $|O P|^{2}=x^{2}+y^{2}=\left(1+k^{2}\right) x^{2}=\frac{a^{2} b^{2}\left(1+k^{2}\right)}{b^{2}+a^{2} k^{2}}$.
Similarly, we have
$$
\begin{array}{l}
|O Q|^{2}=\frac{a^{2} b^{2}\left[1+\left(\frac{m}{k}\right)^{2}\right]}{b^{2}+a^{2}\left(\frac{m}{k}\right)^{2}}=\frac{\left(k^{2}+m^{2}\right) a^{2} b^{2}}{a^{2} m^{2}+b^{2} k^{2}} . \\
\text { Hence } p=|O P|^{2}+|O Q|^{2} \\
=\frac{a^{2} b^{2}\left(1+k^{2}\right)}{b^{2}+a^{2} k^{2}}+\frac{\left(k^{2}+m^{2}\right) a^{2} b^{2}}{a^{2} m^{2}+b^{2} k^{2}} \\
=a^{2} b^{2} \cdot \frac{\left(1+k^{2}\right)\left(a^{2} m^{2}+b^{2} k^{2}\right)+\left(k^{2}+m^{2}\right)\left(b^{2}+a^{2} k^{2}\right)}{\left(b^{2}+a^{2} k^{2}\right)\left(a^{2} m^{2}+b^{2} k^{2}\right)} \\
=a^{2} b^{2} \cdot \frac{\left(a^{2}+b^{2}\right) k^{4}+2\left(a^{2} m^{2}+b^{2}\right) k^{2}+\left(a^{2}+b^{2}\right) m^{2}}{a^{2} b^{2} k^{4}+\left(b^{4}+a^{4} m^{2}\right) k^{2}+a^{2} b^{2} m^{2}} .
\end{array}
$$

Since the value of $p$ is independent of $k$, we have
$$
\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{2\left(a^{2} m^{2}+b^{2}\right)}{b^{4}+a^{4} m^{2}}=\frac{\left(a^{2}+b^{2}\right) m^{2}}{a^{2} b^{2} m^{2}} \text {. }
$$

Thus, we get $m^{2}=\frac{b^{4}}{a^{4}}$, i.e., $m= \pm \frac{b^{2}}{a^{2}}$.
Therefore, $p=a^{2} b^{2} \cdot \frac{a^{2}+b^{2}}{a^{2} b^{2}}=a^{2}+b^{2}$.
(2) Since $m= \pm \frac{b^{2}}{a^{2}}$, the equations of the asymptotes of the hyperbola $\Gamma$ are $y= \pm \frac{b^{2}}{a^{2}} x$.
According to the problem, we can set the equation of the hyperbola $\Gamma$ as $\frac{x^{2}}{a^{4}}-\frac{y^{2}}{b^{4}}=\lambda(\lambda>0)$.
Then $e_{2}=\frac{\sqrt{a^{4} \lambda+b^{4} \lambda}}{\sqrt{a^{4} \lambda}}=\frac{\sqrt{a^{4}+b^{4}}}{a^{2}}$,
$$
e_{1}=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a} \text {. }
$$

Thus, $e_{2}-e_{1}=\frac{\sqrt{a^{4}+b^{4}}}{a^{2}}-\frac{\sqrt{a^{2}}-b^{2}}{a}$

Let $x=\left(\frac{b}{a}\right)^{2}$. Since $a>b>0$, then $0<x<1$.
Let $f(x)=\sqrt{1+x^{2}}-\sqrt{1-x}, x \in[0,1]$.
Since the function $f_{1}(x)=\sqrt{1+x^{2}}$ is increasing on $[0,1]$, and $f_{2}(x)=\sqrt{1-x}$ is decreasing on $[0,1]$, the function $f(x)=f_{1}(x)-f_{2}(x)$ is increasing on $[0,1]$.
Therefore, when $0<x<1$, $f(0)<f(x)<f(1)$, i.e., $0<f(x)<\sqrt{2}$.
Hence, $0<e_{2}-e_{1}<\sqrt{2}$.