3.264 .

$$
\frac{\cos \left(2 x+\frac{\pi}{2}\right) \sin \left(\frac{3 \pi}{2}-3 x\right)-\cos (2 x-5 \pi) \cos \left(3 x+\frac{3 \pi}{2}\right)}{\sin \left(\frac{5 \pi}{2}-x\right) \cos 4 x+\sin x \cos \left(\frac{5 \pi}{2}+4 x\right)}
$$
Solution.

$$
\begin{aligned}
& \frac{\cos \left(2 x+\frac{\pi}{2}\right) \sin \left(\frac{3 \pi}{2}-3 x\right)-\cos (2 x-5 \pi) \cos \left(3 x+\frac{3 \pi}{2}\right)}{\sin \left(\frac{5 \pi}{2}-x\right) \cos 4 x+\sin x \cos \left(\frac{5 \pi}{2}+4 x\right)}= \\
& =\frac{\cos \left(\frac{\pi}{2}+2 x\right) \sin \left(\frac{3 \pi}{2}-3 x\right)-\cos (5 \pi-2 x) \cos \left(\frac{3 \pi}{2}+3 x\right)}{\sin \left(\frac{5 \pi}{2}-x\right) \cos 4 x+\sin x \cos \left(\frac{5 \pi}{2}+4 x\right)}= \\
& =\frac{-\sin 2 x(-\cos 3 x)+\cos 2 x \sin 3 x}{\cos x \cos 4 x-\sin x \sin 4 x}=\frac{\sin 2 x \cos 3 x+\cos 2 x \sin 3 x}{\cos x \cos 4 x-\sin x \sin 4 x}= \\
& =[\sin \alpha \cos \beta+\cos \alpha \sin \beta=\sin (\alpha+\beta) ; \\
& \cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta)]= \\
& =\frac{\sin 5 x}{\cos 5 x}=\operatorname{tg} 5 x .
\end{aligned}
$$

Answer: $\operatorname{tg} 5 x$.