Answer
Please see the work below.
Work Step by Step
(a) We can find the current through each resistor as follows:
$I_{1.5}=\frac{9.0V}{1.5\Omega}$
$\implies I_{1.5}=6.0A$
The current through $2.5\Omega$ is
$I_{2.5}=\frac{9.0V}{2.5\Omega}=3.6A$
We know that
$\frac{1}{R_{eq}}=(\frac{1}{4.8\Omega}+\frac{1}{3.3\Omega}+\frac{1}{8.1\Omega})$
$\implies R_{eq}=1.57\Omega$
$\implies R=R_{eq}+6.3\Omega$
$\implies R=7.88\Omega$
The current thorough $6.3\Omega$ is
$I_{6.3}=\frac{9.0V}{7.88\Omega}$
$\implies I_{6.3}=1.1A$
As $V_p=9.0V-(1.14A)(6.3\Omega)$
$V_p=1.82V$
The current through $4.8\Omega$ is
$I_{4.8}=\frac{1.82V}{4.8\Omega}=0.38A$
Through $3.3\Omega$ $I_{3.3}=\frac{1.82V}{3.3\Omega}=0.55A$
The current through $8.1\Omega$ $I_{8.1}=\frac{1.82V}{8.1\Omega}=0.22A$
(b) We know that the potential difference across the resistor $6.3\Omega$ is given as
$V_{6.3}=(6.3\Omega)(I_{6.3})$
$V_{6.3}=(6.3\Omega)(1.4A)=7.182A$
The potential difference across $1.5\Omega$ is $9.0V$, and the potential difference across $6.3\Omega$ is less than the potential difference across the resistor $1.5\Omega$ because some voltage drops across the parallel combinations.
