Answer
(a) greater
(b) $I_{45\Omega}=0.27A, I_{35\Omega}=I_{82\Omega}=0.103A$
Work Step by Step
(a) We know that $R_{eq}=35\Omega+82\Omega=117\Omega$. We also know that according to Ohm's law, the current through the resistor is inversely proportional to the resistance of the resistor. The equivalent resistance of the upper two resistors in series is greater than the lower resistor. Thus, the current through the $45\Omega$ resistor is greater than the current through the $35\Omega$ resistor.
(b) We know that the net current flowing through the $45\Omega$ resistor is
$I_{45}=\frac{V}{R}$
$I_{45}=\frac{12V}{45\Omega}=0.27A$
Current through $R_{35}$ is
$I_{35}=\frac{V}{R_{eq}}$
$I_{35}=\frac{12V}{117\Omega}=0.103A$
As $35\Omega$ and $82\Omega$ are connected in series, thus the current through these two resistors is the same.
