Because my book doesn't have solutions to these problems, I'm checking here if I solved them correctly (I know it's all probably wrong):
1) $$\tan(\pi+\frac{x}{3})>0$$ What I noticed first is that I could convert tangens to $\frac{\sin}{\cos}$, then remove $\pi$ from both so I finished with: $$tan(\frac{x}{3})>0$$ $$x>3\pi+n\pi, \space n\in \Bbb Z$$ I'm now not quite sure how to interpret this. What I think the best thing to do here is to consider this as interval:$(3\pi+n\pi, +\infty)$, and since the border of $tgx>0$ is $\frac{3\pi}{2}$, the $+\infty$should be replaced by it. And that should be the solution.
2) $$2 \cos(\pi-2x)>1$$ -removing $\pi$ $$\cos(2x)<-\frac{1}{2}$$ It's obvious now that the condition is $\frac{2\pi}{3}<2x<\frac{4\pi}{3}$, if I divide the whole "condition" by 2, I get the interval of the solution:$\frac{\pi}{3}<x<\frac{2\pi}{3}$. (?) I don't know how to use this mechanism in the other problems since then I get different results.
3) $$\cot(\frac{3\pi}{2}-\frac{x}{2})\le\sqrt{3}$$ $$\frac{3\pi}{2}-\frac{x}{2}\le\frac{\pi}{6}+n\pi$$ $$x\geq\frac{2\pi}{3}-2n\pi$$ ... I don't know. This is a dead end.
As the tangent ratio is positive in the first & in the third quadrant.
$$\tan\frac x3>0\implies n\pi<\frac x3<n\pi+\frac\pi2$$ where $n$ is any integer
For the last, $$\cot\left(3\cdot\frac\pi2-\frac x2\right)=\cot\left(\pi+\frac\pi2-\frac x2\right)=\cot\left(\frac\pi2-\frac x2\right)=\tan\frac x2$$
So, we need $\displaystyle\tan\frac x2\le\sqrt3=\tan\frac\pi3$
Now, $\displaystyle\tan\frac x2>\sqrt3$ if $\displaystyle m\pi+\frac\pi3<\frac x2\le m\pi+\frac\pi2$ where $m$ is any integer
We need $U-\{m\pi+\frac\pi3<\frac x2\le m\pi+\frac\pi2\}$ where $U$ is teh Universal Set