Answer
(a) $13.5 \mathrm{~\Omega}$
(b) $17.7 \,\text{A}$
(c) $I_{1} = 5.7 \,\text{A}$ and $I_{2} = 12 \,\text{A}$
Work Step by Step
(a) The equivalent resistance of $R_{1}$ and $R_{2}$ which are in parallel is calculated by
\begin{align*}
R_{\text{eq}} = \dfrac{R_{1}R_{2}}{R_{1} + R_{2} } = \dfrac{(42 \mathrm{~\Omega})(20 \mathrm{~\Omega})}{42 \mathrm{~\Omega} + 20 \mathrm{~\Omega} } =\boxed{13.5 \mathrm{~\Omega}}
\end{align*}
(b) The voltage $V$ across resistors connected in parallel is the same and equals the potential difference across their combination. We use Ohm's law in the next form to get the current
\begin{align*}
I_{t} = \dfrac{V}{R_{\text{eq}}} = \dfrac{ 240 \,\text{V}}{13.5\mathrm{~\Omega}} = \boxed{17.7 \,\text{A}}
\end{align*}
(c) We mentioned in part (b) that the voltage is the same across the two resistors, so again we use Ohm's law to get the current in each resistor by
\begin{align*}
I_{1} = \dfrac{V}{R_{1}} = \dfrac{ 240 \,\text{V}}{42 \mathrm{~\Omega}} = \boxed{5.7 \,\text{A}}
\end{align*}
and,
\begin{align*}
I_{2} = \dfrac{V}{R_{2}} = \dfrac{ 240 \,\text{V}}{20 \mathrm{~\Omega}} = \boxed{12 \,\text{A}}
\end{align*}
