(1) Let the slope of a tangent be $k_{1}$, then the tangent line is
$$\begin{aligned} y & =k_{1} x+\sqrt{k_{1}^{2} a^{2}+b^{2}} \\ \text { or } y & =k_{1} x-\sqrt{k_{1}^{2} a^{2}+b^{2}}, \end{aligned}$$

i.e., $\pm \sqrt{k_{1}^{2} a^{2}+b^{2}}=y-k_{1} x$. Squaring both sides, we get:
$$a^{2} k_{1}^{2}+b^{2}=y^{2}-2 x y k_{1}+k_{1}^{2} x^{2},$$

i.e., $\left(a^{2}-x^{2}\right) k_{1}^{2}+2 x y k_{1}+\left(b^{2}-y^{2}\right)=0$.
Similarly, if the slope of another tangent is $k_{2}$, then we must have:
$$\left(a^{2}-x^{2}\right) k_{2}^{2}+2 x y k_{2}+\left(b^{2}-y^{2}\right)=0.$$

From (A) and (B), we know that $k_{1}, k_{2}$ are the two roots of the equation $\left(a^{2}-x^{2}\right) k^{2}+2 x y k+\left(b^{2}-y^{2}\right)=0$,
thus $k_{1}+k_{2}=\frac{-2 x y}{a^{2}-x^{2}}$.
Given that $\theta_{1}+\operatorname{tg} \theta_{2}=m$,
$$\therefore \frac{-2 x y}{a^{2}-x^{2}}=m,$$

i.e., $m\left(x^{2}-a^{2}\right)=2 x y$ is the required equation.
(2) From $\left(a^{2}-x^{2}\right) k^{2}+2 x y k+\left(b^{2}-y^{2}\right)=0$,

we get $k_{1}+k_{2}=\frac{-2 x y}{a^{2}-x^{2}}, k_{1} k_{2}=\frac{b^{2}-y^{2}}{a^{2}-x^{2}}$.
Given the condition $\operatorname{ctg} \theta_{1}+\operatorname{ctg} \theta_{2}=n$,
$$\begin{array}{l} \text { i.e., } \frac{1}{k_{1}}+\frac{1}{k_{2}}=n, \frac{k_{1}+k_{2}}{k_{1} k_{2}}=n, \\ \therefore \frac{-2 x y}{b^{2}-y^{2}}=n, \text { i.e., } n\left(y^{2}-b^{2}\right)=2 x y \end{array}$$

is the required equation.