(7) Let the function $f(x)=\left\{\begin{array}{l}\frac{1}{p}\left(x=\frac{q}{p}\right), \\ 0\left(x \neq \frac{q}{p}\right),\end{array}\right.$ where $p, q$ are coprime (prime), and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$, and $f(x)>\frac{1}{5}$ is $\qquad$
(7) 5 Hint: Clearly, $x=\frac{q}{p}$ (otherwise $f(x)=0$). At this point, from $f(x)=\frac{1}{p}=\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$.

When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$. Therefore, there are 5 values of $x$ that satisfy the condition.