1. Function $f(x)=x^{a}, x \in(-1,0) \cup(0$, 1). If the inequality $f(x)>|x|$ holds, then under the condition $\alpha \in$ $\left\{-2,-1,-\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3}, 1,2\right\}$, the number of values $\alpha$ can take is ( ).
(A) 4
(B) 3
(C) 2
(D) 1
$-1 . \mathrm{A}$.
Let $I=(0,1) \cup(-1,0)$. To make $f(x)=x^{a}>|x|$, $x^{a}$ should be greater than 0 on $I$, knowing that $a=-1, 1, \frac{1}{3}$ obviously do not meet the requirements.

When $a=0$, $f(x)=1,0<|x|$ holds;
When $\alpha=2$, on $I$, $f(x)>1,0<|x|$;

When $a=\frac{2}{3}$, $f(x)=|x|^{\frac{2}{3}}$, by $\frac{|x|}{f(x)}=|x|^{\frac{1}{3}}$ and $0<|x|$;

When $\alpha=-\frac{2}{3}$, $f(x)=|x|^{-\frac{2}{3}}$, by $\frac{|x|}{f(x)}=|x|^{\frac{5}{3}}$ and $0<|x|$.
In summary, there are 4 values that $a$ can take.