Find the number of elements in each factorgroup. 
(a) and (c) are 2 and 3 and I don't think I have a problem with those.
However for (b) I get 11 cosets but they are not disjoint. According to theory there should only be 6. So what is happening here?

The image of $8\mathbb Z$ in $\mathbb Z/12\mathbb Z$ is the same as the image of $\langle 8,12\rangle=\gcd(8,12)\mathbb Z=4\mathbb Z$.
This same idea works for all items.
The precise idea hint at above is to consider the canonical projection $\pi:\mathbb Z \to \mathbb Z/12\mathbb Z$. Then $\pi^{-1}(\langle 8\rangle)=\gcd(8,12)\mathbb Z=4\mathbb Z$ and so $\dfrac{\mathbb Z/12\mathbb Z}{\langle 8\rangle} \cong \dfrac{\mathbb Z}{4\mathbb Z}$.
