#### Answer

$i_{1}=i_{2}=-0.5A$

#### Work Step by Step

This question references the same circuit as that in the previous question (3.7). We are now expected to use mesh current analysis to obtain $i_{1}$ and $i_{2}$.
By KCL, $\sum{i_{n}}=0$. Taking KCL at the node immediately to the left of $R_{2}$, $1A$ of current enters the node and $i_{1}A$ exits it, leaving $0+1-i_{1}=(1-i_{1})A$ of current to pass through $R_{2}$.
Taking KCL at the node above $R_{3}$,
$$(1-i_{1})-i_{2}-2=0$$
$$\Rightarrow{i_{1}}+i_{2}=-1$$
By considering current flow in the leftmost and rightmost loops of the circuit, we can deduce that $3i_{1}=1$ and $6i_{2}=2$. (This is known as the loop current method).
By solving the above two equations and taking ratios in $i_{1}$ and $i_{2}$, we can deduce that $i_{1}=i_{2}$.
Substituting $i_{1}=i_{2}$ into the equation we derived earlier,
$$i_{2}+i_{2}=2i_{2}=-1\Rightarrow{i_{2}}=-0.5A$$
$$\therefore{i_{1}}=i_{2}=-0.5A$$