Let S8 be the symmetric group of degree 8.
1 2 3 4 5 6 7 8
6 8 3 7 1 4 5 2
Write down the elements of the cyclic subgroups in S8 generated by α and by α^(−1) respectively.
So I found a^(-1) to be
12345678
58367142
and a in disjoint form is (16475)(28)
a^(-1) in disjoint form is (15746)(28)
what do I do now?
First of all any element and its inverse (in any group) generate the same subgroup, this easily follows from the fact that any subgroup which contains $\alpha$ must contain $\alpha^{-1}$ and vice versa. So you don't need to think of this as two different exercises.
Anyway, the group which is generated by one element is $\langle\alpha\rangle=\{\alpha^k:k\in\mathbb{Z}\}$. If the order of $\alpha$ is finite and equal to $n$ then it is equal to $\{\alpha^k: 0\leq k\leq n-1\}$. So just start multiplying $\alpha$ by itself until you get the identity. This way you will get the whole subgroup $\langle\alpha\rangle$.