3.069. $\frac{1-\cos (8 \alpha-3 \pi)}{\tan 2 \alpha-\cot 2 \alpha}$.
Solution.

$$
\begin{aligned}
& \frac{1-\cos (8 \alpha-3 \pi)}{\tan 2 \alpha-\cot 2 \alpha}=\frac{1-\cos (3 \pi-8 \alpha)}{\frac{\sin 2 \alpha}{\cos 2 \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}=\frac{(1-\cos (3 \pi-8 \alpha)) \sin 2 \alpha \cos 2 \alpha}{\sin ^{2} 2 \alpha-\cos ^{2} 2 \alpha}= \\
& =-\frac{(1-\cos (3 \pi-8 \alpha)) \cdot 2 \sin 2 \alpha \cos 2 \alpha}{2\left(\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha\right)}=-\frac{(1+\cos 8 \alpha) \sin 4 \alpha}{2 \cos 4 \alpha}= \\
& =-\frac{\left(1+2 \cos ^{2} 4 \alpha-1\right) \sin 4 \alpha}{2 \cos 4 \alpha}=-\frac{2 \cos ^{2} 4 \alpha \sin 4 \alpha}{2 \cos 4 \alpha}=\frac{-2 \sin 4 \alpha \cos 4 \alpha}{2}= \\
& =-\frac{\sin 8 \alpha}{2}
\end{aligned}
$$

Answer: $-\frac{\sin 8 \alpha}{2}$.