It should be obvious that the rotations form a normal subgroup of index 2 (the fact that it is of index 2 is sufficient to prove normality). If we call the rotation subgroup $R$, and the reflection coset $F$, we have:
$RR = R$
$RF = F$
$FR = F$
$FF = R$.
Subgroups containing only rotations are cyclic, due to the fact that $R$ is cyclic. We thus get exactly one subgroup of order $d$ contained in $R$, for each divisor $d$ of $n$.You should prove any (and all) of these subgroups are normal.
Subgroups containing only reflections and the identity must have order a power of 2. Since a reflection times a reflection is a rotation, with $(r^ks)(r^ms) = r^{k-m}$, it should be clear that any such subgroup is in fact of order 2 (we must have $k = m)$. These subgroups are typically NOT normal, but there is an exception for $n = 2$ ($D_2 = V$ is abelian).
Which brings us to "mixed subgroups", containing at least one rotation, and one reflection. These are going to be of the form $\langle r^k,s\rangle$ where $k|n$, that is, isomorphic to $D_m$, where $m = \dfrac{n}{k}$ . You can think of these as symmetries of an $m$-gon, which are also symmetries of an $n$-gon, since $n$ is a multiple of $m$ (the axes of symmetry of an $n$-gon include all the axes of symmetry of the $m$-gon, plus more).
These "mixed subgroups" aren't, in general, normal, but in some special cases they are: for example if $n$ is even and $k = \dfrac{n}{2}$, or $k = 2$.
Another case worth mentioning is when $n$ is an odd prime; in this case, any subgroup containing more than one reflection, or a reflection and a (non-trivial) rotation, is the entire group, which limits the possibilities for subgroups.
For a more complete analysis, see: https://in.answers.yahoo.com/question/index?qid=20091014113730AA7KJDt
$<a^3>$ is $\{a^3, a^6, a, a^4, a^7, a^2, a^5, 1\} = C_8$. So clearly, $a^6$ and $a^4$ belong to the same coset. Same with the second problem.
$<a^3b> = \{a^3b, 1\}$ because $a^3ba^3b = a^3a^{-1}ba^2b = a^2ba^2b = abab = b^2 = 1$. The cosets are $\left\{\{a^3b, 1\}, \{a^4b, a\}, \{a^5b, a^2\}, \{a^6b, a^3\}, \{a^7b, a^4\}, \{b, a^5\}, \{ab, a^6\}, \{a^2b, a^7\}\right\}$.
Best Answer
${D}_n \not\cong \mathbb{Z}/2 \oplus \mathbb{Z}/n,$ as the latter is Abelian and the former is not.