[Solution] From $\log _{a} \frac{1}{b}>\log _{a} \sqrt{b}$, we get $-\log _{a} b>\frac{1}{2} \log _{a} b$, so $\log _{a} b<0$. If $0<a<1$, then $b<1$ or $a>1, b>\sqrt{b}$, which implies $b>1$. From $\sqrt{b}>\log _{b} a^{2}$, we get $\frac{1}{2} \log _{a} b>2 \log _{b} a=\frac{2}{\log _{a} b}$,
and since $\log _{a} b<0$, we have $\left(\log _{a} b\right)^{2}<4$, which means $-2<\log _{a} b<0$.
Since $\log _{a} b$ is an integer, we have $\log _{a} b=-1$, i.e., $b=\frac{1}{a}$, thus $a b=1$. Conclusion (4) is correct.
Therefore, the answer is $(A)$.
9.10 If the decimal part of $\log _{a} b$ is 0, and $\log _{a} \frac{1}{b}>\log _{a} \sqrt{b}>\log _{b} a^{2}$, then among the following four conclusions:
(1) $\frac{1}{b}>\sqrt{b}>a^{2}$.
(2) $\log _{a} b+\log _{b} a=0$.
(3) $0<a<b<1$.
(4) $a b-1=0$.

the number of correct conclusions is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(China Junior High School Mathematics League, 1990)