Definition: Let $G$ be a group. A subnormal series for G is a chain of subgroups $1 = G_0 \subseteq G_1 \subseteq G_2 \subseteq G_n = G$ such that $G_i$ is normal in $G_{i+1}$ for $i = 0,1, ..., (n-1).$
We call a group soluble if it has a subnormal series and $G_{i+1}/G_i$ is abelian for $i = 0,1, ..., (n-1).$
First of all, i've been told the following lemma is useful.
Let $G$ be a group and $N$ a normal subgroup. Then $G/N$ is abelian if and only if $[g,h] \in N$ for all $g, h \in G$.
I want to prove points (ii), (iii) and (iv) of the following (I have included point (i), which used the above lemma, in case it helps see what kind of path I should be taking):
(i) Subgroups of soluble groups are soluble.
(ii) If $\phi: G \rightarrow H$ is a homomorphism and G is soluble then Im($\phi$) is soluble.
(iii) Quotient groups of soluble groups are soluble (i.e. if $N$ is a normal subgroup of $G$ and $G$ is soluble then $G/N$ is soluble).
(iv) If $N$ is a normal subgroup of $G$ and both $N$ and $G/N$ are soluble then $G$ is soluble.
First, note that ii and iii are equivalent since $Im(\phi)\cong G/Ker(\phi)$. To prove this, just take the subnormal series for $G$ and project it onto $G/N$, which should preserve normality and abelianness of consecutive quotients (you can check that the commutator of $G_{i+1}/N$ is contained in $G_i/N$).
For iv, since subgroups of $G/N$ correspond to subgroups of $G$ containing $N$, you can build a series from $1$ to $N$, and then use the series from $1$ to $G/N$ to get a series from $N$ to $G$, which you can combine with the other one to get a series from $1$ to $G$.