3. Find the values of the following expressions:
(1) $\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$;
(2) $\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \sin 20^{\circ} \cdot \cos 80^{\circ}$;
(3) $\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right)$;
(4) $\cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{5 \pi}{15} \cdot \cos \frac{6 \pi}{15} \cdot \cos \frac{7 \pi}{15}$.
3. (1) Let $A=\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}, B=\cos 10^{\circ} \cdot \cos 30^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}$, then $A \cdot B=$ $\frac{1}{16} \sin 20^{\circ} \cdot \sin 60^{\circ} \cdot \sin 100^{\circ} \cdot \sin 140^{\circ}=\frac{1}{16} \cos 10^{\circ} \cdot \cos 30^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}=\frac{1}{16} B$, and $B \neq 0$, so $A=$ $\frac{1}{16}$, that is, $\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}=\frac{1}{16}$.
Or by $\sin 3 \theta=4 \sin \theta \cdot \sin \left(60^{\circ}-\theta\right) \cdot \sin \left(60^{\circ}+\theta\right)$ to find, also get $\frac{1}{16}$.
(2) Let $A=\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \sin 20^{\circ} \cdot \cos 80^{\circ}, B=\cos ^{2} 20^{\circ}+\sin ^{2} 80^{\circ}-\sqrt{3} \cos 20^{\circ} \cdot \sin 80^{\circ}$, then $A+$ $B=2-\sqrt{3}\left(\cos 20^{\circ} \cdot \sin 80^{\circ}-\sin 20^{\circ} \cdot \cos 80^{\circ}\right)=2-\sqrt{3} \sin 60^{\circ}=\frac{1}{2}, B-A=\cos 40^{\circ}-\cos 160^{\circ}-$ $\sqrt{3}\left(\cos 20^{\circ} \cdot \sin 80^{\circ}+\sin 20^{\circ} \cdot \cos 80^{\circ}\right)=2 \cos 30^{\circ} \cdot \cos 10^{\circ}-\sqrt{3} \sin 100^{\circ}=0$. Thus $2 A=\frac{1}{2}, A=\frac{1}{4}$, that is, $\sin ^{2} 20^{\circ}+\cos ^{2} 80^{\circ}+\sqrt{3} \cdot \sin 20^{\circ} \cdot \cos 80^{\circ}=\frac{1}{4}$.
(3) Let $m=\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right), n=\sin ^{2} A+\sin ^{2}\left(60^{\circ}-A\right)+\sin ^{2}\left(60^{\circ}+A\right)$, then $x+y=3, y-x=\cos 2 A+\cos 2\left(60^{\circ}-A\right)+\cos 2\left(60^{\circ}+A\right)=\cos 2 A+2 \cos 120^{\circ} \cdot \cos 2 A=0$. Thus $y=x=$ $\frac{3}{2}$, that is, $\cos ^{2} A+\cos ^{2}\left(60^{\circ}-A\right)+\cos ^{2}\left(60^{\circ}+A\right)=\frac{3}{2}$.
(4) Let $M=\cos \frac{\pi}{15} \cdot \cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{5 \pi}{15} \cdot \cos \frac{6 \pi}{15} \cdot \cos \frac{7 \pi}{15}, N=\sin \frac{\pi}{15} \cdot \sin \frac{2 \pi}{15}$. $\sin \frac{3 \pi}{15} \cdot \sin \frac{4 \pi}{15} \cdot \sin \frac{5 \pi}{15} \cdot \sin \frac{6 \pi}{15} \cdot \sin \frac{7 \pi}{15}$, then $2^{7} M \cdot N=\sin \frac{2 \pi}{15} \cdot \sin \frac{4 \pi}{15} \cdot \sin \frac{6 \pi}{15} \cdot \sin \frac{8 \pi}{15} \cdot \sin \frac{10 \pi}{15} \cdot$ $\sin \frac{12 \pi}{15} \cdot \sin \frac{14 \pi}{15}=N$, and $N \neq 0$, so $M=\frac{1}{2^{7}}$, that is, the original expression $=\frac{1}{128}$.