$$
\begin{array}{l}
\frac{\sin ^{2} 30^{\circ}+\sin ^{2} 355^{\circ}+\sin ^{2} 40^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 50^{\circ}+\sin ^{2} 55^{\circ}+\sin ^{2} 60^{\circ}}{\tan 36^{\circ} \cdot \tan ^{3} 39^{\circ} \cdot \tan ^{5} 42^{\circ} \cdot \tan ^{7} 45^{\circ} \cdot \tan ^{5} 48^{\circ} \cdot \tan ^{3} 55^{\circ} \cdot \tan 4^{\circ}} . \\
= \\
\hline
\end{array}
$$
2, 1.3.5.
Notice,
$$
\sin ^{2} \alpha+\sin ^{2}\left(90^{\circ}-\alpha\right)=\sin ^{2} \alpha+\cos ^{2} \alpha=1 \text {, }
$$

and $\sin ^{2} 45^{\circ}=\frac{1}{2}$.
When $n$ is a positive integer,
$$
\tan ^{n} \alpha \cdot \tan ^{n}\left(90^{\circ}-\alpha\right)=\tan ^{n} \alpha \cdot \cot ^{n} \alpha=1 \text {, }
$$

and $\tan 45^{\circ}=1$.
Then the original expression $=3.5$.