A $7$-Sylow subgroup of $G$ has order $7$; the number of $7$-Sylow subgroups must be congruent to $1$ modulo $7$, and must divide $140 = 7\times 5\times 4$. Thus, it must divide $20$ and be congruent to $1$ modulo $7$; the only possibility is that the number of $7$-Sylow subgroups of $G$ is one (hence it is normal; in fact, characteristic) Thus, the subgroup $H$ of order $35$ must contain the unique $7$-Sylow subgroup of $G$.
Likewise, a $5$-Sylow subgroup of $G$ has order $5$, and the number of $5$-Sylow subgroups of $G$ must be congruent to $1$ modulo $5$ and divide $140$, hence must divide $7\times 4$. The only possibility is that there is a unique $5$-Sylow subgroup of $G$ (which is therefore characteristic, and thus normal), which must also be contained in $H$.
Thus, $H$ must be product of the unique $7$-Sylow subgroup and the unique $5$-Sylow subgroup of $G$. Since $H$ is the product of characteristic subgroups, it is in fact characteristic and therefore normal in $G$.
(Note that any group of order 35 is necessarily cyclic, by the by).