Answer
a). There would be no change in $I_{1}, I_{2}, I_{4}$. In the lower branch, the current will be halved since the new resistance is now doubled to 4Ohm.
b). $I_{1} = I_{2}=3A$ and $I_{4}=6A$
Originally, $I_{3}=6A$ but now the new currents are $I'_{3}=I'_{5}=3A$
Work Step by Step
$R_{1}+R_{2}$, $R_{4}+R_{5}$, and $R_{3}$ are in parallel.
Therefore $\frac{1}{Re}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}$
$Re=1 Ohm$
$i=12/1=12A$
Since potential across $R_{4}+R_{5}$ & $R_{3}$ are same,
$i_{1}\times (R_{4}+R_{5})=(i-2i_{1})\times R_{3}$
$i_{1}=12/4=3A$
So, current in the upper arm = $3A$ and that in lower arm = $6A$
Originally, $I_{3}=6A$ but now the new currents are $I'_{3}=I'_{5}=3A$
