= Solution
For
$$
g(z)=\frac{e^z-1}{z\log(1+z)},
$$
both $e^z-1$ and $\log(1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its <residue> is
$$
\operatorname{Res}(g,0)
=\lim_{z\to0}\frac{e^z-1}{\log(1+z)}
=\boxed{1}.
$$
For $h(z)=\sin z\sin(1/z)$, multiplication of the two Laurent series shows that all powers are even:
$$
h(z)=
\sum_{p,q\geq0}
\frac{(-1)^{p+q}}{(2p+1)!(2q+1)!}\,
z^{2(p-q)}.
$$
There are infinitely many negative powers, so zero is an <essential singularity>. There is no $z^{-1}$ term, and therefore
$$
\boxed{\operatorname{Res}(h,0)=0}.
$$
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