= Solution
The apparent <singularity> at zero cancels: the <Taylor series> of the <sine> gives
$$
\frac{\sin z}{z^2}-\frac1z=-\frac z6+\frac{z^3}{120}-\cdots.
$$
Thus zero is a <removable singularity>, with extension $f(0)=1/2$. At $z=-2$, the other terms are <holomorphic>, whereas $(z+2)^{-1}$ has a <simple pole>. Therefore
$$
\boxed{z=0:\ \text{removable},\qquad z=-2:\ \text{simple pole with residue }1.}
$$
There are no other finite <singularities>. If “all” includes the point at infinity on the <Riemann sphere>, it is an <essential singularity>: $f(1/w)=w/(1+2w)+w^2\sin(1/w)-w$ has infinitely many negative powers in its <Laurent series> at $w=0$.
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