4. (11th "Hope Cup" Invitational Competition Question) If the even function $f(x)$ defined on $\mathbf{R}$ is increasing on $[0,+\infty)$, and $f\left(\frac{1}{3}\right)=0$, then the solution to the inequality $f\left(\log _{\frac{1}{8}} x\right)>0$ is ( ).
A. $\left(\frac{1}{2}, 1\right)$
B. $(2,+\infty)$
C. $\left(0, \frac{1}{2}\right) \cup(2,+\infty)$
D. $\left(\frac{1}{2}, 1\right) \cup(2,+\infty)$
4. C. Reason: Since $f(x)$ is an even function and $f\left(\frac{1}{3}\right)=0$, it follows that $f\left(-\frac{1}{3}\right)=0$. Also, since $f(x)$ is increasing on $[0,+\infty)$, it is decreasing on $(-\infty, 0]$.
From $f\left(\log _{\frac{1}{8}} x\right)>0$, we know $\log _{\frac{1}{8}} x\frac{1}{3}$, solving this gives $x>2$ or $0<x<\frac{1}{2}$.