The value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$? I want to get the value of $I=\int_{|z|=2}z^2\sin\frac{1}{z}dz$.
My idea is to let $t=\frac{1}{z}$, then 
$$I=\int_{|z|=2}z^2\sin\frac{1}{z}dz=\int_{|t|=\frac{1}{2}}\frac{\sin t}{t^4}dt=2\pi i \times-\frac{1}{6}=-\frac{\pi i}{3}$$ Is this right?

Your result has the wrong sign because you did not take into account that the substitution $t=-\frac{1}{z}$ reverts the orientation of the circle. 
You could also apply the residue theorem directly to
$$
 z^2 \sin \frac 1z  = z^2 \left( \frac 1z - \frac{1}{3!z^3} + \ldots \right) = z -\frac{1}{6z} + \ldots
$$
so that 
$$\operatorname{Res}(z^2 \sin \frac 1z, 0) = -\frac 16$$ and therefore
$$
I = 2 \pi i \operatorname{Res}(z^2 \sin \frac 1z, 0) = - \frac{\pi i}{3} \,.
$$
