15. Given $f(x)=\log _{a} \frac{1-m x}{x-1}$ is an odd function $(a>1, a \neq 1)$.
(1) Find the value of $m$;
(2) Based on the result of (1), determine the monotonicity of $f(x)$ on $(1,+\infty)$ (no proof required);
(3) If when $x \in(r, a-2)$, the range of $f(x)$ is exactly $(1,+\infty)$, find the values of $a$ and $r$.
15. (1) From $f(-x)=-f(x)$ we get $m=-1$ (discard $m=1$); (2) $f(x)=\log _{a} \frac{1+x}{x-1}$. Let $u=$ $\frac{x+1}{x-1}=1+\frac{2}{x-1}, u$ is decreasing on $(1,+\infty)$, so when $a>1$, $f(x)$ is decreasing on $(1,+\infty)$. When $0<a<1$, $f(x)$ is increasing on $(1,+\infty)$. Solving $\frac{1+x}{x-1}>0$, we get $x<-1$ or $x>1$, so the domain of $f(x)$ is $(-\infty,-1) \cup(1,+\infty)$, and the range is $(-\infty, 0) \cup(0,+\infty)$. (1) When $(r, a-2) \subset(-\infty,-1)$, $a-2 \leqslant-1, a \leqslant 1$, so $0<a \leqslant 1$, and $r<a-2 \leqslant -1$, which is not possible. (2) When $(r, a-2) \subset(1,+\infty)$, $r \geqslant 1, a>3, f(x)$ is decreasing on $(r, a-2)$, then $f(a-2)=1, \log _{a} \frac{a-1}{a-3}=1, a=2+\sqrt{3}, r=1$.