Using the residue theorem, first find the poles, $4z^2-1=0$, $z=\pm. \frac{1}{2}$. Both poles are inside the unit circle, so:
$Res_{z=\frac{1}{2}}(\frac{z^2sinz}{4z^2-1})=\frac{1}{16}sin\frac{1}{2}$
$Res_{z=\frac{-1}{2}}(\frac{z^2sinz}{4z^2-1})=\frac{1}{16}sin\frac{1}{2}$
$2\pi i(\frac{1}{16}sin\frac{1}{2}+\frac{1}{16}sin\frac{1}{2})=\frac{\pi i}{4}sin{\frac{1}{2}}$
Looking for any feedback... Thanks.