1. Among $\frac{1}{7}, \frac{\pi}{2}, 0.2002, \frac{1}{2}(\sqrt{3-2 \sqrt{2}}-\sqrt{2})$, $\frac{\sqrt{n}-\sqrt{n-2}}{3}(n(n>3)$ is an integer $)$, the number of fractions is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
-1 (B).
$$
\frac{1}{2}(\sqrt{3-2 \sqrt{2}}-\sqrt{2})=\frac{1}{2}(\sqrt{2}-1-\sqrt{2})=-\frac{1}{2} \text {, }
$$

When $n(n>3)$ is an integer, one of $\sqrt{n}$ and $\sqrt{n-2}$ is irrational, meaning that $n$ and $n-2$ cannot both be perfect squares at the same time. Let $n=s^{2}, n-2=t^{2}$. Then we have $s^{2}-t^{2}=2, (s+t)(s-t)=2 \times 1, s+t=2, s-t=1$. Since $s=\frac{3}{2}, t=\frac{1}{2}$ are not integer solutions, $\frac{\sqrt{n}-\sqrt{n-2}}{3}$ is not a fraction. Therefore, there are 3 fractions: $\frac{1}{7}, 0.2002, \frac{1}{2}(\sqrt{3-2 \sqrt{2}}-\sqrt{2})$.