Isomorphism theorem Is it true the following result? 
Let $f:G \mapsto G^\prime$ be a surjective morphism of groups, and let $H$ be a normal subgroup in $G$. Then $f(H)$ is a normal subgroup in $G^\prime$ (this part is easy to be checked)  and $G/ H \backsimeq G^\prime /f(H)$. (?)
 A: The second part holds if kernel of $f$, denoted $\ker(f)$, is contained in the normal subgroup $H$. Proving this is a good exercise. Supposing that the kernel is not contained in $H$ will give you lots of counter-examples. For example, consider the following counter-example.
Take $G=\mathbb{Z}_4\times \mathbb{Z}=\langle a, b\rangle$ where $a$ has order four and $b$ has order two. Take $H=\langle b\rangle$ and take $f$ to be the natural homomorphism from $G$ to $\mathbb{Z}_2\times\mathbb{Z}_2$. This has kernel $\ker(f)=\langle a^2\rangle$ (note that $H\cap\ker(f)$ is trivial). Then quotienting out by the image of $\langle b\rangle$, that is, by $f(H)$, we obtain $\mathbb{Z}_2$. However, if we has quotiented out by $\langle b\rangle$ originally we would have obtained the cyclic group of order four, $\mathbb{Z}_4$.
Another counter-example would be to take $G=H\times K$ and consider the image of $K$ in $G/H$.
A: The answer is yes to the first part, the image via a surjective homomorphism of a normal subgroup is normal, but the second in not true in general.
To prove the first part you can either prove it directly. 
If $H \lhd G$ is a normal subgroup then for every $h' \in f(H)$ and for every $g' \in G'$ you have to prove that $g' h' g'^{-1} \in f(H)$. From the surjectivity there are $h \in H$ and $g \in G$ such that $f(g)=g'$ and $f(h)=h'$ and so $f(ghg^{-1})=g' h' g'^{-1}$ and by normality of $H$ we have that $ghg^{-1} \in H$, so $g'h'g'^{-1} \in f(H)$.
Another approach to prove this is use the general fact that 
$$f^{-1}(f(H))=H \ker f =\{ xy \mid x \in H, y \in \ker f\} $$
Since both $H$ and $\ker f$ are normal for every $x \in H,y \in \ker f$ and for every $g \in G$ we have that 
$$gxyg^{-1}=gxg^{-1}gyg^{-1} \in H \ker f$$
and so also this subgroup is normal.
Because $f$ is surjective homomorphism $G'$ is a quotient of $G$ and so by the theorems of quotients the image of a subgroup in the quotient give a one-on-one correspondence between normal subgroups of $G'$ and those subgroups of $G$ that contains $\ker f$. From this general theorem and the fact that $H \ker f$ is normal it follows that $f(H \ker f) =f(H)$ is normal.
About the last part of the question here's a counterexample:
consider the group $G = (\mathbb Z/2 \mathbb Z)^2$ and let $f \colon (\mathbb Z/2 \mathbb Z)^2 \to\mathbb Z/2 \mathbb Z=G'$ be the projection on the first component. Then $\ker f=\{0\} \times \mathbb Z/2 \mathbb Z$ and the subgroup $\mathbb Z/2 \mathbb Z \times \{0\}$ has image the whole quotient group $\mathbb Z/2 \mathbb Z$. So $G'/f(H)=0$ while $G/H \cong \mathbb Z/2\mathbb Z$.
