At $z=0$ the function $f(z)=\exp({z\over 1-\cos z})$ has

How do you know it isn't a pole? For small $z$, $\frac{z}{1-\cos z}\approx \frac{z}{z^2}=\frac{1}{z}$, so that for small $z$ your function is $\displaystyle e^{\frac{1}{z}}=\Sigma \frac{1}{n!z^n}$, and this clearly goes to infinity faster than any polynomial goes to $0$, ie. there is no $m$ such that $\,\,\displaystyle z^m\Sigma \frac{1}{n!z^n}$ is bounded near $0$, hence it is an essential singularity.

$\frac{z}{1-cos z}=\frac{z}{(z^2/2!)-(z^4/4!)+...}=\frac{1}{(z/2)+O(z^3)}=\frac{2}{z} [\frac{1}{1+O(z^2)}]=\frac{2}{z} ([{1+O(z^2)})^{-1} =\frac{2}{z} [{1-O(z^2)}]=\frac{2}{z}-O(z)$ Thus $$exp [\frac{z}{1-cos z}]=exp[\frac{2}{z}-O(z)]=1+\frac{\frac{2}{z}-O(z)}{1!}+\frac{(\frac{2}{z}-O(z))^2}{2!}+... $$ Obviously the above Laurent series expansion of f(z) valid around z=0 i.e. for $0<\vert{z}\vert<\epsilon$ has infinitely many negative (and positive) powers of z. Consequently, z=0 is an essential singularity of f(z).
The correct options are (3) and (4).