10. If the odd function $f(x)$ is defined on $(-\infty, 0) \cup(0,+\infty)$, and is increasing on $(0,+\infty)$, with $f(1)=0$, and there is a function $g(\theta)=\sin ^{2} \theta+m \cos \theta-2 m, \theta \in\left[0, \frac{\pi}{2}\right]$. If the set $M=\{m \mid$ $g(\theta)<0\}$, and the set $N=\{m \mid f[g(\theta)]<0\}$, find the set $M \cap N$.
10. Given that $f(x)$ is an odd function on $(-\infty, 0) \cup(0,+\infty)$, and it is an increasing function on $(0,+\infty)$, with $f(1)=0$, it is easy to see that $f(-1)=-f(1)=0$, and $f(x)$ is an increasing function on $(-\infty, 0)$.
Therefore, from $f(x)\sin ^{2} \theta+1$,
we have $m>\frac{2-\cos ^{2} \theta}{2-\cos \theta}=\frac{4-\cos ^{2} \theta-2}{2-\cos \theta}=2+\cos \theta-\frac{2}{2-\cos \theta}=4-\left[(2-\cos \theta)+\frac{2}{2-\cos \theta}\right]$.
And $-\left[(2-\cos \theta)+\frac{2}{2-\cos \theta}\right] \leqslant-2 \sqrt{2}$, with equality holding if and only if $2-\cos \theta=\frac{2}{2-\cos \theta}$, i.e., $\cos \theta=2-\sqrt{2}$.
Therefore, $m>4-2 \sqrt{2}$.
Thus, $M \cap N=\{m \mid m>4-2 \sqrt{2}\}$.