3. $\left(2 \sin 36^{\circ} \cdot 2 \sin 72^{\circ}\right)^{2}=$
3.5.
$$
\begin{array}{l}
\text { Original expression }=4 \times 2 \sin ^{2} 36^{\circ} \times 2 \sin ^{2} 72^{\circ} \\
=4\left(1-\cos 72^{\circ}\right)\left(1-\cos 144^{\circ}\right) .
\end{array}
$$

And $\cos 72^{\circ}+\cos 144^{\circ}$
$$
\begin{array}{l}
=\frac{1}{2 \sin 36^{\circ}}\left(2 \sin 36^{\circ} \cdot \cos 72^{\circ}+\right. \\
2 \sin 36^{\circ} \cdot \cos 144^{\circ} \text { ) } \\
=\frac{1}{2 \sin 36^{\circ}}\left(\sin 108^{\circ}-\sin 36^{\circ}+\right. \\
\sin 180^{\circ}-\sin 108^{\circ} \text { ) } \\
=\frac{1}{2 \sin 36^{\circ}}\left(-\sin 36^{\circ}\right)=-\frac{1}{2} \text {, } \\
\cos 72^{\circ} \cdot \cos 144^{\circ} \\
=\frac{1}{4 \sin 72^{\circ}}\left(2 \sin 144^{\circ} \cdot \cos 144^{\circ}\right) \\
=\frac{1}{4 \sin 72^{\circ}} \sin 288^{\circ} \\
=\frac{1}{4 \sin 72^{\circ}}\left(-\sin 72^{\circ}\right)=-\frac{1}{4} \text {, } \\
\end{array}
$$

Therefore, the original expression
$$
\begin{array}{l}
=4\left(1-\cos 72^{\circ}-\cos 144^{\circ}+\cos 72^{\circ} \cdot \cos 144^{\circ}\right) \\
=4\left(1+\frac{1}{2}-\frac{1}{4}\right)=5 .
\end{array}
$$