Answer
a) 1 A
b) 0 A
c) 3 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 side only and hence it is given by
$$I_{\rm battery}=\dfrac{\Delta V_{\rm battery }}{R_{\rm eq}}$$
The battery current passes through the two resistors since they are in series.
Plug the known;
$$I_{\rm battery}=\dfrac{ 30}{ 30}=\color{red}{\bf 1}\;\rm A$$
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. This makes the current passes through the 10-$\Omega$ resistor and no current passes through the 20-$\Omega$ resistor.
In other words, the inductor shorts out the 20-$\Omega$ resistor.
$$I_{\rm battery}=\dfrac{\Delta V_{\rm battery }}{R_{\rm 10\Omega}}=\dfrac{30}{10}=\bf 10\;\rm A$$
$$I_{\rm 20\Omega} =\color{red}{\bf 0}\;\rm A$$
$$\color{blue}{\bf [c]}$$
Immediately after opening the switch, the current will move through the $LR$ circuit. According to the conservation of current, at the moment we open the switch the current is 3 A, as we found above, But that will not remain for a long time since it will decay with time.
Hence, immediately after opening the switch, the current that moves through the 20-$\Omega$ is
$$I_{\rm 20\Omega} =\color{red}{\bf 3}\;\rm A$$
