(2) If the function $f(x)=\log _{a} x(a>0$, and $a \neq 1)$ satisfies $f\left(\frac{2}{a}\right)>f\left(\frac{3}{a}\right)$, then the solution set of $f\left(1-\frac{1}{x}\right)>1$ is ( ).
(A) $0<x<\frac{1}{a}$
(B) $0<x<\frac{1}{1-a}$
(C) $1<x<\frac{1}{a}$
(D) $1<x<\frac{1}{1-a}$
(2) From
$$
f\left(\frac{2}{a}\right)=\log _{a} \frac{2}{a}>f\left(\frac{3}{a}\right)=\log _{a} \frac{3}{a},
$$

we get $0<a<1$. Therefore, the original inequality is equivalent to the inequality
$$
0<1-\frac{1}{x}<a \text {. }
$$

Solving this inequality yields $1<x<\frac{1}{1-a}$. Hence, the answer is D.