DC LR circuit with current source
For part (b), the circuit to the left of the switch is disconnected so you can disregard it for any \$t&gt;0\$.

The key insight for (b) is that the circuit is once again in DC steady state (all transients decay as \$t \rightarrow \infty\$).

Thus, the voltage across the inductor is \$0V\$, i.e., it can be replaced with an ideal wire.

Now, this leaves a simple current divider circuit so the inductor current is given by current division.