New Year Enigma Contd

Here I shall provide a relatively elementary approach (in comparison to @Otto Bretscher 's!):

$\sin\frac{\pi}{14}\cdot \sin\frac{3\pi}{14}\cdot \sin\frac{5\pi}{14}$ $=\cos\frac{6\pi}{14}\cdot \cos\frac{4\pi}{14}\cdot \cos\frac{2\pi}{14}$ $=\cos\frac{\pi}{7}\cdot \cos\frac{2\pi}{7}\cdot \cos\frac{3\pi}{7}$ $=-\cos\frac{\pi}{7}\cdot \cos\frac{2\pi}{7}\cdot \cos\frac{4\pi}{7}$ $=\frac{-\sin\frac{\pi}{7}\cdot \cos\frac{\pi}{7}\cdot \cos\frac{2\pi}{7}\cdot \cos\frac{4\pi}{7}}{\sin\frac{\pi}{7}}$ $=\frac{-\sin\frac{2\pi}{7}\cdot \cos\frac{2\pi}{7}\cdot \cos\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}$ $=\frac{-\sin\frac{4\pi}{7}\cdot \cos\frac{4\pi}{7}}{4\sin\frac{\pi}{7}}$ $=\frac{-\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}$ $=\frac{1}{8}\cdot \frac{\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}}$ $=\boxed{0.125}.$

Nice variant of Morrie's Law

Using $\cos(x)=\sin(\pi/2-x)$ and $\prod_{k=0}^{n-1}\cos(2^kx)=\frac{\sin(2^nx)}{2^n\sin(x)}$ we can write the given expression as $-\cos(8\pi/14)\cos(4\pi/14)\cos(2\pi/14)=-\frac{\sin(16\pi/14)}{2^3\sin(2\pi/14)}=\boxed{0.125}$