= Solution
At the origin, the <Taylor series> $\sin z=z+O(z^3)$ and $1-\cos z=z^2/2+O(z^4)$ give
$$
f_1(z)=z^{-3}(1+O(z)),\qquad f_2(z)=2z^{-4}(1+O(z^2)).
$$
Thus \b[$f_1$ has a pole of order three and $f_2$ a pole of order four at zero]. For the third function, the <Laurent series>
$$
f_3(z)=\sum_{k=0}^{\infty}\frac{(-1)^k z^{1-2k}}{(2k+1)!}
$$
has infinitely many nonzero negative-power terms, so \b[zero is an essential singularity of $f_3$].
To classify the point at infinity on the <Riemann sphere>, use the coordinate $w=1/z$. Then
$$
f_3(1/w)=w^{-2}\sin w=w^{-1}-\frac w6+O(w^3),
$$
so \b[$f_3$ has a simple pole at infinity].
For the other two functions, an important qualification is necessary. The <poles> of $f_1$ at nonzero integer multiples of $\pi$, and those of $f_2$ at nonzero integer multiples of $2\pi$, approach infinity. Equivalently, their images in the $w$ coordinate accumulate at zero. Hence \b[infinity is a non-isolated singularity of $f_1$ and $f_2$]. Neither function is <holomorphic> on any punctured neighbourhood of infinity, so the <classification of isolated singularities> into removable singularities, poles and essential singularities does not apply there.
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