358. $3 \sin \frac{\pi}{3}-2 \cos \frac{\pi}{6}+3 \tan \frac{\pi}{3}-4 \cot \frac{\pi}{2}$.
Solution. From the table, we take the values $\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} ; \cos \frac{\pi}{6}=$ $=\frac{\sqrt{3}}{2} ; \operatorname{tg} \frac{\pi}{3}=\sqrt{3}, \operatorname{ctg} \frac{\pi}{2}=0$ and substitute them into the given expression:

$$
3 \cdot \frac{\sqrt{3}}{2}-2 \cdot \frac{\sqrt{3}}{2}+3 \sqrt{3}-4 \cdot 0=\frac{\sqrt{3}}{2}+3 \sqrt{3}=\frac{7 \sqrt{3}}{2}
$$