3.244. $\sin ^{2} 2 \alpha+\sin ^{2} \beta+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)$.
## Solution.

$$
\begin{aligned}
& \sin ^{2} 2 \alpha+\sin ^{2} \beta+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =\frac{1-\cos 4 \alpha}{2}+\frac{1-\cos 2 \beta}{2}+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =\frac{1}{2}(1-\cos 4 \alpha+1-\cos 2 \beta)+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =-\left(\frac{\cos 4 \alpha}{2}+\frac{\cos 2 \beta}{2}-1\right)+\cos (2 \alpha+\beta) \cos (2 \alpha-\beta)= \\
& =\left[\cos x \cos y=\frac{1}{2}(\cos (x-y)+\cos (x+y))\right]= \\
& =-\frac{\cos 4 \alpha}{2}-\frac{\cos 2 \beta}{2}+1+\frac{\cos 4 \alpha}{2}+\frac{\cos 2 \beta}{2}=1
\end{aligned}
$$

Answer: 1 .