Five. (This question is worth 20 points) Given an odd function $f(x)$ defined on $(-\infty, 0) \cup(0,+\infty)$, and it is an increasing function on $(0, +\infty)$, with $f(1)=0$. Also, the function
$g(\theta)=\sin ^{2} \theta+m \cos \theta-2 m, \theta \in\left[0, \frac{\pi}{2}\right]$, and the sets
$M, N$ satisfy $M=\{m \mid g(\theta)>0\}, N=\{m \mid$ $f[g(\theta)]<0\}$. Find $M \cap N$.
Five, since the odd function $f(x)$ is increasing on $(0,+\infty)$, then $f(x)$ is also increasing on $(-\infty, 0)$.
Furthermore, because $f(1)=0$, we have
$$
f(-1)=-f(1)=0 .
$$

Therefore, simultaneously satisfying $g(\theta)0 .
\end{array}
$$

Let $\cos \theta=t$, then $t \in[0,1]$. Suppose
$$
\varphi(t)=t^{2}-m t+2 m-2 .
$$

Then this problem is equivalent to: find $m$ such that $\varphi(t)>0, t \in$ $[0,1]$. This is equivalent to finding the value of $m$ when the minimum value of the function $\varphi(t)$ for $t \leqslant[0,1]$ is greater than 0. Consider the position of the axis of symmetry $t=\frac{m}{2}$ relative to $[0,1]$.
(1) When $\frac{m}{2}0 .
\end{array}\right.
$$

There is no solution in this case.
(2) When $0 \leqslant \frac{m}{2} \leqslant 1$, $\varphi\left(\frac{\pi}{2}\right)$ is the minimum value. Solve
$$
\left\{\begin{array}{l}
0 \leqslant \frac{m}{2} \leqslant 1, \\
\varphi\left(\frac{m}{2}\right)=\frac{-m^{2}+8 m-8}{4}>0
\end{array}\right.
$$

We get $\quad 4-2 \sqrt{2}1$, $\varphi(1)$ is the minimum value. Solve
$$
\left\{\begin{array}{l}
\frac{m}{2}>1, \\
\varphi(1)=m-1>0
\end{array}\right.
$$

We get $m>2$.
Combining (1), (2), and (3), we get $m>4-2 \sqrt{2}$.