 Solution

For the first function, the complex exponential has no zeros, while the denominator has three distinct simple zeros. Thus there are $\boxed{\text{simple poles at }z=0,2i,-2i;\text{no finite zeros}}$. To examine infinity, substitute $z=1/w$:The factor $e^{1/w}$ has an essential singularity at zero, and multiplication by a nonzero finite-order factor cannot remove it. Therefore $\boxed{\infty \text{ is an isolated essential singularity}}$ of the first function.

For the second function, $\sin z$ has simple zeros at $z=n\pi$. At zero, its Taylor series givesso $\boxed{z=0\text{ is removable, with extended value }1}$. Every other $z=n\pi$ is a double pole, since the numerator is nonzero there. There are no finite zeros, including after removal at zero.

At infinity the double poles $n\pi$ accumulate. Equivalently, after $z=1/w$, the function $1/[w^2{\sin }^2(1/w)]$ has double poles at $w=1/(n\pi )$ tending to zero. Hence $\boxed{\infty \text{ is a non-isolated singularity}}$. The removable/pole/essential trichotomy for isolated singularities does not apply there; calling this an isolated essential singularity would be incorrect.