1. $\star \star$ Given the ellipse $\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m>n>0)$ passing through the fixed point $P(a, b)(a, b \in \mathbf{R}$, and $a b \neq 0,|a| \neq|b|)$, find the minimum value of $m+n$.
From the given, we have $\frac{a^{2}}{m^{2}}+\frac{b^{2}}{n^{2}}=1$, which can be rewritten as $\left(\frac{a}{m}\right)^{2}+\left(\frac{b}{n}\right)^{2}=1$. Also, $a b \neq 0$. By symmetry, we can assume without loss of generality that $a, b \in \mathbf{R}^{+}$. Therefore, we can set $\left\{\begin{array}{l}\frac{a}{m}=\cos \alpha, \\ \frac{b}{n}=\sin \alpha,\end{array} \quad \alpha \in\left(0, \frac{\pi}{2}\right)\right.$, hence $\left\{\begin{array}{l}m=\frac{a}{\cos \alpha}, \\ n=\frac{b}{\sin \alpha} .\end{array}\right.$ Therefore,
$$
\begin{aligned}
(m+n)^{2} & =\left(\frac{a}{\cos \alpha}+\frac{b}{\sin \alpha}\right)^{2}=a^{2} \sec ^{2} \alpha+b^{2} \csc ^{2} \alpha+\frac{2 a b}{\sin \alpha \cdot \cos \alpha} \\
& =a^{2}\left(\tan ^{2} \alpha+1\right)+b^{2}\left(\cot ^{2} \alpha+1\right)+\frac{2 a b\left(1+\tan ^{2} \alpha\right)}{\tan \alpha}
\end{aligned}
$$

Let $t=\tan \alpha, t>0$. Then we have
$$
\begin{aligned}
(m+n)^{2} & =a^{2}\left(t^{2}+1\right)+b^{2}\left(\frac{1}{t^{2}}+1\right)+\frac{2 a b\left(t^{2}+1\right)}{t} \\
& =a^{2} t^{2}+\frac{b^{2}}{t^{2}}+2 a b t+\frac{2 a b}{t}+a^{2}+b^{2} \\
& =a\left(a t^{2}+\frac{b}{t}+\frac{b}{t}\right)+b\left(\frac{b}{t^{2}}+a t+a t\right)+a^{2}+b^{2} \\
& \geqslant a \cdot 3 \sqrt[3]{a t^{2} \cdot \frac{b}{t} \cdot \frac{b}{t}}+b \cdot 3 \sqrt[3]{\frac{b}{t^{2}} \cdot a t \cdot a t}+a^{2}+b^{2} \\
& =3 a^{\frac{4}{3}} b^{\frac{2}{3}}+3 a^{\frac{2}{3}} b^{\frac{4}{3}}+a^{2}+b^{2}=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{3}
\end{aligned}
$$

Equality holds if and only if $\left\{\begin{array}{l}a t^{2}=\frac{b}{t}, \\ \frac{b}{t^{2}}=a t,\end{array}\right.$ which is when $t=\sqrt[3]{\frac{b}{a}}$. Therefore, the minimum value of $(m+n)^{2}$ is $\left(a^{\frac{2}{3}}+\right.$
$$
\left.b^{\frac{2}{3}}\right)^{3} \text {, hence }(m+n)_{\min }=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}} \text {. }
$$