Let $f(z) := \frac{3\sin 4z}{z^4}$. Find $$\mbox{res}\left [f(z),\infty\right ]$$ For $a=\infty$, the residue is given as $-\frac{1}{2i\pi}\int_{|z|=R} f(z)\mbox{d}z$. $$\int_{|z|=R} \frac{3\sin 4\xi}{(\xi-0)^{3+1}}\mbox{d}\xi = \frac{2i\pi}{3!}g^{(3)}(0),\ \ g(z) := 3\sin 4z $$ We find that $g^{(3)}(0) = -192\cos 0 =-192$, so we have: $$\mbox{res}\left [f(z),\infty\right ] = 32 $$
An alternative attempt:
Let $w=1/z$, Find the Laurent series of $f(1/w)$ around point $1/a=0$
$$f(1/w) = 3w^4\sin (4/w) = 3w^4\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\cdot\frac{4^{2k+1}}{w^{2k+1}}= 3\sum_{k=0}^\infty \frac{(-1)^k4^{2k+1}}{(2k+1)!w^{2k-3}},\ |w|>0$$
Substituting back gives us
$$f(z) = 3\sum_{k=0}^\infty \frac{(-1)^k4^{2k+1}}{(2k+1)!}z^{2k-3},\ \ |z|<\infty\,(?) $$
Desired residue at $k=1$ as required.
You inverted $w$ to $z$ twice. It's either $z^{2k-3}$ or $1/z^{3-2k}$. If you correct that, you get the right residue $32$ for $k=1$ (where you need to take the negative value of the coefficient).