$1 \Omega$ and $3 \Omega$ are in series so,
$egin{aligned}
& R_s=R_1+R_2 \\
& R_s=3+1=4 \Omega
\end{aligned}$
and $R_2=8 \Omega$
net $i=$ current in the circuit
Current through $R_1$ is $i_1=\frac{i \times R_2}{R_1+R_2}$
$=\frac{i \times 8}{12}=\frac{21}{3}$
Current $R_2$ is $i_2=\frac{1 \times R_1}{R_1+R_2}=\frac{i \times 4}{12}$
$=\frac{i \times 8}{12}$
Power dissipated in $3 \Omega$ resistance is: $P_1=l_1^2 \times 3$
Power dissipated in $8 \Omega$ resistance is: $P_2=l_2^2 \times 8$
For (i) and (ii)
$\begin{aligned}
& \frac{P_1}{P_2}=\frac{i_1^2 \times 3}{i_2^2 \times 8} \\
& \Rightarrow \frac{P_1}{P_2}
\end{aligned}$
$\begin{aligned}
& =\frac{\left(\frac{2 i}{3}\right)^2 \times 3}{\left(\frac{i}{3}\right)^2 \times 8}=\frac{3}{2} \\
\therefore \quad P_1 & =\frac{3}{2} \times P_2=\frac{3}{2} \times 2 =3 \mathrm{~W}
\end{aligned}$
$\begin{aligned}
\because & P & =100 \mathrm{~W}, V=125 V \\
\therefore & P & =V I \\
\Rightarrow & I & =\frac{P}{V}=\frac{100}{125} A
\end{aligned}$