51. Given the function $f(x)=\frac{x^{2}+c}{a x+b}$ is an odd function, $f(1) < f(3)$, and the solution set of the inequality $0 \leqslant f(x) \leqslant \frac{3}{2}$ is $[-2,-1] \cup [2,4]$.
(1) Find $a, b, c$;
(2) Does there exist a real number $m$ such that the inequality $f(-2+\sin \theta) < -m^{2}+\frac{3}{2}$ holds for all $\theta \in \mathbb{R}$? If it exists, find the range of $m$; if not, explain the reason.
51. (1) $f(x)$ is an odd function $\Leftrightarrow f(-x)=f(x)$ for all $x$ in the domain $\Leftrightarrow b=0$, thus $f(x)=\frac{1}{a}\left(x+\frac{c}{x}\right)$. Also, $\left\{\begin{array}{l}f(2) \geqslant 0, \\ f(-2) \geqslant 0\end{array} \Leftrightarrow\left\{\begin{array}{l}f(2) \geqslant 0, \\ -f(2) \geqslant 0\end{array} \Leftrightarrow f(2)=0 \Leftrightarrow c=-4\right.\right.$, and from $f(1)0, \\ c3\end{array}\right.$. Thus, we determine $a>0$.

At this point, $f(x)=\frac{1}{a}\left(x-\frac{4}{x}\right)$ is an increasing function on $[2,4]$, noting that $f(2)=0$, then it must be that $f(4)=\frac{3}{2}$, i.e., $\frac{1}{a}\left(4-\frac{4}{4}\right)=\frac{3}{2}$,
$\therefore a=2$, in summary, $a=2, b=0, c=-4$.
(2) From (1), $f(x)=\frac{1}{2}\left(x-\frac{4}{x}\right)$, which is an increasing function on $(-\infty, 0)$ and $(0,+\infty)$, and $-3 \leqslant-2+\sin \theta \leqslant-1$, so the range of $f(-2+\sin \theta)$ is $\left[-\frac{5}{6}, \frac{3}{2}\right]$,

The real number $m$ that meets the condition should satisfy $\frac{3}{2}-m^{2}>\frac{3}{2}$, i.e., $m^{2}<0$, hence there is no real number $m$ that meets the condition.