Explanation

(d)
(1)We have$,\sin{A = \sin B}$
$\Rightarrow A = B\ or\ A = \pi - B$
Now$,\sin{2A =}\sin{2B\ }$is satisfied by $A = B$ but it is not satisfied by $A = \pi - B$
(2)$\cos{\frac{\pi}{7}\cos{\frac{4\pi}{7}\cos{\frac{5\pi}{7} = \cos\frac{\pi}{7}\cos{\frac{4\pi}{7}\cos\left( \pi - \frac{2\pi}{7} \right)}}}}$
$= - \cos{\frac{\pi}{7}\cos\frac{2\pi}{7}}\cos\frac{4\pi}{7}$
$= - \frac{\sin\left( 2^{3}\frac{\pi}{7} \right)}{2^{3}\sin\frac{\pi}{7}} = - \frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}} = \frac{1}{8}$