$$
\begin{aligned}
& \text { If } A=\left\{x \in[0,2 \pi] / \tan x-\tan ^2 x>0\right\} \\
& \therefore \quad \tan x-\tan ^2 x>0 \\
& \Rightarrow \tan x(1-\tan x)>0 \\
& \quad 0 < \tan x < 1
\end{aligned}
$$
So, $\quad 0 < x < \frac{\pi}{4}$ and $\pi < x < \frac{5 \pi}{4}$
$$
\Rightarrow \quad x \in\left(0, \frac{\pi}{4}\right) \cup\left(\pi, \frac{5 \pi}{4}\right)
$$
Now, $B=\left\{x \in[0,2 \pi] /|\sin x| < \frac{1}{2}\right.$
$$
\begin{aligned}
& \Rightarrow \quad-\frac{1}{2} < \sin x < \frac{1}{2} \\
& \quad \sin x < \frac{1}{2} \text { and } \sin x>\frac{-1}{2} \\
& \Rightarrow \quad 0 < x < \frac{\pi}{6} \text { and }\left(\frac{5 \pi}{6}, 2 \pi\right) \text { and } x \in\left(0, \frac{7 \pi}{6}\right) \\
& \Rightarrow \quad x \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\end{aligned}
$$
So, $A \cap B$ is defined in
$$
\left(0, \frac{\pi}{6}\right) \cup\left(\pi, \frac{7 \pi}{6}\right)
$$