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?

Consider [b]. The element $8$ in $\mathbb{Z}/12$ has order $12/\gcd(8,12) = 3$: indeed we have $8+8+8 = 24 \equiv 0 \mod 12$. Thus the order of $(\mathbb{Z}/12)/\langle 8 \rangle$ is $12/3 = 4$. Explicitly, the cosets are $\langle 8 \rangle = \{0,8,4\}$, $1+\langle 8 \rangle = \{1,9,5\}$, $2+\langle 8\rangle = \{2,10,6\}$ and $3+\langle 8 \rangle = \{3,11,7\}$.
More generally, the order of $n$ in $\mathbb{Z}/m$ is $m/\gcd(n,m)$ and so the order of the factor group $(\mathbb{Z}/m)/\langle n \rangle$ is $m/(m/\gcd(n,m)) = \gcd(n,m)$.
