Image of subgroup and Kernel of homomorphism form subgroups
Solution 1:
It's ok but don't forget to show that $\mathrm{Im}(H)\ne \emptyset$ since $f(e)=e'\in\mathrm{Im}(H)$ and the same thing for $\ker f$.
Solution 2:
Yes, your proofs are correct. To be perfectly rigorous you also need to say that these sets are nonempty (which is obvious, but still).