The residue of a function $f$ represented by a Laurent Series in complex analysis is defined as the coefficient at $n=-1$ of the series ($a_{-1}$). Giving the definition of the cossine in complex analysis:
$$f(z)=\cos(\frac{1}{z})=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n}$$ So the series is only defined from $n=0 $ to $\infty$.
Is the residue $0$ at $z=0$ because the series of the function is undefined at $n=-1$? Or is it because the function only has even exponents of $z$ (and so there is no term $z^{-1}$).
And, finally, is the residue at a singularity $\operatorname{res}(z_0, f)=0$ for every function that can only be expanded as a series from $n=0\to \infty$? (I'm still a bit confused on this whole topic.)
Before getting to the residue, we need to correct your series expansion of the cosine. It should be
$\cos(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}$.
Notice that the exponent of $z$ is $2n$, not $n$. Now we can insert $1/z$, and use $(1/z)^n=z^{-n}$:
$\cos(1/z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(1/z)^{2n}=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{-2n}$.
But to find the residue, we have to find the coefficient $a_{-1}$ of the Laurent series of the form $\sum_{n=-\infty}^\infty a_n z^n$. The exponent of $z$ here is $n$, not $-2n$. So we have to find the term in the series above where $z^{-1}$ appears. This term does not exist, since the exponent is always even. This means that its coefficient is $0$, and so is the residue.