Three. (20 points) For the function $f(x)$, if there exists $x_{0} \in \mathbf{R}$ such that $f\left(x_{0}\right)=x_{0}$, then the point $\left(x_{0}, x_{0}\right)$ is called a fixed point of the function $f(x)$.
(1) If the function $f(x)=a x^{2}+b x-b(a \neq 0)$ has fixed points $(1,1)$ and $(-3,-3)$, find the values of $a$ and $b$;
(2) If for any real number $b$, the function $f(x)=a x^{2} + b x - b$ always has 2 distinct fixed points, find the range of real values for $a$;
(3) If the odd function $g(x)$ defined on the real number set $\mathbf{R}$ has (a finite number of) $n$ fixed points, prove that $n$ must be an odd number.
(1) By the definition of a fixed point, we have $f(x)-x=0$, that is,
$$
a x^{2}+(b-1) x-b=0 \text {. }
$$

Substituting $x=1$ into equation (1) yields $a=1$.
Substituting $x=-3$ into equation (1) yields
$$
9 a-3(b-1)-b=0 \text {. }
$$

Solving this gives $b=3$.
(2) According to the condition, for any real number $b$, equation (1) always has two distinct real roots, so $\Delta=(b-1)^{2}+4 a b>0$ always holds, which means for any real number $b, b^{2}+(4 a-2) b+1>0$ always holds.
Thus, $\Delta^{\prime}=(4 a-2)^{2}-4<0$, that is,
$16 a(a-1)<0$.
Solving this gives $0<a<1$.
(3) Clearly, the point $(0,0)$ is a fixed point of the odd function $g(x)$ on $\mathbf{R}$. If $g(x)$ has a fixed point $\left(x_{0}, x_{0}\right)$ different from $(0,0)$, then $g\left(x_{0}\right)=x_{0}$.

Also, $g\left(-x_{0}\right)=-g\left(x_{0}\right)=-x_{0}$, so $\left(-x_{0},-x_{0}\right)$ is also a fixed point of $g(x)$ on $\mathbf{R}$.

Therefore, except for the origin, the finite fixed points of $g(x)$ all appear in pairs, with $2 k$ of them $(k \in \mathbf{N})$. Thus, $g(x)$ has a total of $2 k$ +1 fixed points on $\mathbf{R}$. Therefore, $n$ is an odd number.