Let $\mathbb{F}$ be a free group on $a$ and $b$. Let $N$ be the normal subgroup of $\mathbb{F}$ generated by $a^2,b^3$ and $(ab)^2$. Similarly $H$ is the normal subgroup generated by $a^2,b^3$ and $aba^{-1}b^{-1}$.
i) Show that $N\neq H$.
ii)Find Schreier representative for $\mathbb{F}$ mod $H$ and $\mathbb{F}$ mod $N$.
On
If I understand you correctly, a Schreier system of representatives for both subgroups should be provided by $$ 1, b, b^{2}, a, b a, b^{2} a $$
Addendum The reason is that both quotient groups are a semidirect product of a cyclic group $\langle b \rangle$ of order three by a cyclic group $\langle a \rangle$ of order $2$. Then one of them is abelian (and thus cyclic of order $6$) while the other is not (and thus isomorphic to the symmetric group on three letters, a.k.a. the dihedral group of order $6$).
I don't know the Schreier representative, however there is some information I can provide.
$$F/N=\langle a,b \mid b^3,a^2,aba^{-1}=b^{-1}\rangle\cong D_6$$ $$F/H=\langle a,b \mid b^3,a^2,ab=ba \rangle\cong C_2\oplus C_3$$