Find the order of the pole and the residue of $f(x)=\frac{\cos z}{z^2}$ and $g(x)=\frac{e^z-1}{z^2}$

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Find the order of the pole and the residue of $$f(z)=\frac{\cos z}{z^2}$$ and $$g(z)=\frac{e^z-1}{z^2}$$

What I said:

$f(z)=\frac{\cos z}{z^2}$ has a pole of order $2$ at $z=0$. Then: $Res(f,0)=-\sin(0)=0$. Can a residue be $0$?

For $g(z)$: Taylor expand $\frac{e^z-1}{z^2}=1/z+1/2!+z/3!+z^2/4!...$ which implies that $1$ is the residue. What would be the order of this?