Answer
a) ${\bf 0.5}\;\rm A$
b) ${\bf 1.0}\;\rm A$
Work Step by Step
$$\color{blue}{\bf [a]}$$
At the instant the switch is closed, the current is assumed to split into two currents, $I_1$ through the left side of the circuit and $I_2$ through the right side of the circuit at which there is an inductor.
But we know that the voltage across the inductor is given by $\Delta V_L=(dI/dt)L$ where at the instant we close the switch $dI/dt$ approaches zero since the change in the current at this instant is infinitesimal.
This means that the current at the instant we close the switch will move through the left resistor and hence it is given by
$$I_{\rm battery}=\dfrac{\Delta V_{\rm battery }}{R_{\rm left}}$$
Plug the known;
$$I_{\rm battery}=\dfrac{ 10}{ 20}=\color{red}{\bf 0.5}\;\rm A$$
Note that we built our answer as we assume that the inductor is an ideal one which means that it has no resistance and that the battery has no internal resistance too.
$$\color{blue}{\bf [b]}$$
After closing the switch for a long time, the current in the circuit reaches a definite value and remains constant.
Hence, $\Delta V_L=(dI/dt)L=0$ V.
Thus this (ideal) inductor now works as an ideal wire with zero resistance. Now the two resistors are in parallel, so the battery current is given by
$$I_{\rm battery}=\dfrac{\Delta V_{\rm battery }}{R_{\rm eq}}=\dfrac{10}{10}$$
$$I_{\rm battery}=\dfrac{ 10}{ 20}=\color{red}{\bf 1.0}\;\rm A$$
