= 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 $z=1/w$:
$$
\frac{e^{1/w}}{[(1/w)^2+4]/w}=\frac{w^3e^{1/w}}{1+4w^2}.
$$
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> gives
$$
\frac{z^2}{\sin^2z}=\left(\frac{z}{z-z^3/6+O(z^5)}\right)^2=1+\frac{z^2}{3}+O(z^4),
$$
so $\boxed{z=0\text{ is removable, with extended value }1}$. 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.
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