Answer
(a) $I_{1} =8 \,\text{A}$ and $I_{3} = 12 \,\text{A}$
(b) $\varepsilon$ = 84 V
Work Step by Step
(a) $R_{1}$ and $R_{2}$ are in parallel, so the voltage is the same for both of them, in this case we could use the value of $V_2=I_{2}R_{2}$ to get the current $I_1$ by
\begin{align*}
I_{1} = \dfrac{V_{2}}{R_{1}} = \dfrac{I_{2}R_{2} }{R_{1}} = \dfrac{( 4.0 \,\text{A})(6.0 \mathrm{~\Omega})}{3.0 \mathrm{~\Omega}} = \boxed{8 \,\text{A}}
\end{align*}
The current is the same for $R_{12}$ and $R_{3}$as they are in series, ($I_{12} = I_{3}$). Therefore the current in $R_3$ equals the sum of $I_{1}$ and $I_{2}$
$$I_{3} = I_{1} + I_{2} = 4.0 \,\text{A} + 8.0 \,\text{A} = \boxed{12 \,\text{A}}$$
(b) The emf of the battery is
\begin{equation*}
\varepsilon= IR_{eq}
\end{equation*}
We want to find the equivalent resistance in the circuit $R_{eq} $. $R_{1}$ and $R_{2}$ are in parallel, so $R_{12}$ is
\begin{align*}
R_{\text{12}} = \dfrac{R_{1}R_{2}}{R_{1} + R_{2} } = \dfrac{(3.0 \mathrm{~\Omega})(6.0 \mathrm{~\Omega})}{3.0 \mathrm{~\Omega} + 6.0 \mathrm{~\Omega} } =2 \mathrm{~\Omega}
\end{align*}
$R_{12}$ is in series with $R_{3}$ so we get the equivalent resistance $R_{\text{eq}}$ in the circuit by
\begin{align*}
R_{\text{eq}} = R_{12} + R_{3} = 2.0 \mathrm{~\Omega} + 5.0 \mathrm{~\Omega} = 7.0 \mathrm{~\Omega}
\end{align*}
Hence, $\varepsilon$ is
\begin{align*}
\varepsilon= IR_{eq} = (12.0 \,\text{A}) (7.0 \mathrm{~\Omega}) = \boxed{84 \,\text{V}}
\end{align*}
