Answer
a) $I_1=\frac{2E}{3R}$ and $I_2=I_3=\frac{E}{3R}$
b) $I_3=0A$ and $I_1=I_2=\frac{E}{2R}$
c) $V_C=\frac{1}{2}E$
Work Step by Step
The voltage of the capacitance is given by $E(1-e^{-t/RC})$
a) At $t=0$, $V_C=0$ so the capacitance is like a wire.
$R_{eq}=\Big(\frac{1}{R}+\frac{1}{R}\Big)^{-1}+R=1.5R$
$I_1=\frac{2E}{3R}=I_2+I_3$
$V_2=V_3$
$I_2R=I_3R$
$I_2=I_3=\frac{E}{3R}$
b) At $t=\infty$, the capacitance is fully charged with a voltage of $E$. Therefore, there will be no potential difference across $R_3$ and $I_3=0A$ and $I_1=I_2=\frac{E}{2R}$
c) $V_C=V_{R2}=I_2R=\frac{1}{2}E$
