Answer
See the detailed answer below.
Work Step by Step
We know that the potential difference across an inductor is given by
$$\Delta V_L=-L\dfrac{dI}{dt}$$
We are given $L$ and $\Delta V$ for both inductors, so we can't find which has a larger current than the other, but we can find $dI/dt$ which means we can find which one has a rapid change in current.
$$\color{blue}{\bf [a]}$$
No.
$$\color{blue}{\bf [b]}$$
Yes, and to find which is which, we need to use the formula above for the two conductors. Solving this formula for $dI/dt$,
$$ \left|\left(\dfrac{dI}{dt}\right)_1\right|=\left|\dfrac{(\Delta V_L)_1}{-L_1}\right| =\left|\dfrac{2}{2}\right|=\bf 1\;\rm A/s$$
$$ \left|\left(\dfrac{dI}{dt}\right)_2\right|=\left|\dfrac{(\Delta V_L)_2}{-L_2}\right| =\left|\dfrac{4}{1}\right|=\bf 4\;\rm A/s$$
Thus, Inductor 2 is the one at which the current is changing more rapidly.
$$\color{blue}{\bf [c]}$$
In both cases, the current is decreasing since $\dfrac{dI}{dt}\lt 0$ and $\Delta V_L\gt 0$