8. The solution set of the inequality $\left|\frac{1}{\frac{2}{\tan x}-1}\right|>2$ is
8. $\left.|x| x \in\left(k \pi+\arctan \frac{4}{3}, k \pi+\arctan 2\right) \cup(k \pi+\arctan 2, k \pi+\arctan 4) k \in Z\right\}$ $\frac{1}{\frac{2}{\tan x}-1}>2$ or $\frac{1}{\frac{2}{\tan x}-1}<-2$, and note that $\tan x \neq 0$.
We get $\tan x \in\left(\frac{4}{3}, 2\right) \cup(2,4)$
Therefore, $x \in\left(k \pi+\arctan \frac{4}{3}, k \pi+\arctan 2\right) \cup(k \pi+\arctan 2, k \pi+\arctan 4) k \in \mathbf{Z}$.