To find right cosets of H in G where G= and H=$$ ,where o(G)=10 To find right cosets of H in G where G= and H=$<a^{2}>$ ,where o(G)=10
Since order of $G =10$ , so $a^{10}=e$ .We have
$G= { a,a^{2},a^{3},a^{4},a^{5},a^{6},a^{7},a^{8},a^{9},e}$
and 
$H = {a^{2} ,a^{4},a^{6},a^{8},e}$
So $Ha={a^{3},a^{5},a^{7},a^{9},e}$
$Ha^{2} = H$
Is this right way to attempt this

Almost. The only difference is that the coset need not be a group. The only change to be made is $Ha=\{a,a^3,a^5,a^7,a^9\}$. Then as you've noticed every element is in either $Ha$ or $H$. It is a nice exercise to prove that being in a coset is an equivalence relation and the that cosets partition the group. In this way each element will be in a coset, and only one coset.
