You found one such cyclic subgroup.
We also have a cyclic subgroup that's also a subgroup of the cyclic group you found, of order two:$$\{1, a^2\}$$
Four additional cyclic subgroups of order two are as follows:
$$\{1, x\}, \{1, b\}, \{1, c\}, \{1, d\}$$
Of course, we need also to add a sixth, trivially cyclic but distinct subgroup: $\{1\}$.
So, with the subgroup you found, and the additional 6 subgroups here, we have, in all, $7$ distinct cyclic subgroups of $D_8$.
If $K\subseteq G$ is a subgroup, then we define $gK=\{gk\mid k\in K\}$ to be a left coset of $K.$
Since $K\subseteq G$ is a subgroup, then the order of $K$ divides the order of $G$ (by Lagrange's theorem). That is to say, the number of distinct cosets determined by $K\subseteq G$ is precisely $\frac{|G|}{|K|}.$ The rest, as you have in your problem statement, is a matter of sitting down and writing out some direct computation. What I have stated about the number of distinct cosets is what tells you've written down all of the cosets you can.
Best Answer
If you know Lagranges theorem then it doesn't take much calculation to know that dihedral groups can be generated by two elements.
I presume you know that $D_8$ has order $8$. There is a very obvious element which has order $4$, call it $\rho$. Take $\tau$ to be any element outside of $\{1, \rho, \rho^2, \rho^3\}$. Then the subgroup generated by $\rho$ and $\tau$ has more than $4$ elements in it (all the powers of $\rho$ plus the element $\tau$) and by Lagranges theorem the order of this subgroup must divide $8$. Well the only number bigger than $4$ that divides $8$ is $8$, so the subgroup generated by $\rho$ and $\tau$ is all of $D_8$.