$8.481 \cdot \frac{1}{4} \tan \frac{x}{4}+\frac{1}{2} \tan \frac{x}{2}+\tan x=2 \sqrt{3}+\frac{1}{4} \cot \frac{x}{4}$.
Solution. $\frac{1}{4}\left(\operatorname{tg} \frac{x}{4}-\operatorname{ctg} \frac{x}{4}\right)+\frac{1}{2} \operatorname{tg} \frac{x}{2}+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow$

$\Leftrightarrow \frac{1}{2} \cdot \frac{\sin ^{2} \frac{x}{4}-\cos ^{2} \frac{x}{4}}{2 \sin \frac{x}{4} \cdot \cos \frac{x}{4}}+\frac{1}{2} \operatorname{tg} \frac{x}{2}+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow$
$\Leftrightarrow-\frac{1}{2} \cdot \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}+\frac{1}{2} \operatorname{tg} \frac{x}{2}+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow \frac{1}{2}\left(\operatorname{tg} \frac{x}{2}-\operatorname{ctg} \frac{x}{2}\right)+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow$

$\Leftrightarrow \frac{\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}+\operatorname{tg} x=2 \sqrt{3} \Leftrightarrow \operatorname{tg} x-\operatorname{ctg} x=2 \sqrt{3} \Leftrightarrow$

$\Leftrightarrow \frac{\sin ^{2} x-\cos ^{2} x}{2 \sin x \cdot \cos x}=\sqrt{3} \Leftrightarrow \operatorname{ctg} 2 x=-\sqrt{3} \Leftrightarrow 2 x=\frac{5 \pi}{6}+\pi k \Rightarrow x=\frac{5 \pi}{12}+\frac{\pi k}{2}$.

Answer: $x=\frac{\pi}{12}(6 k+5)$.