How does current divide in parallel inductors?
Lets start by putting down what we know



The voltage across each inductor must be the same (just like the current must be the same to series capacitors).
Now the voltage induced in an inductor is

$$V = L \frac{\mathrm{d}i}{\mathrm{d}t}$$

These voltages must be equal so that

$$ V_{AB}
= L_1 \frac{\mathrm{d}(i_1)}{\mathrm{d}t}
= L_2 \frac{\mathrm{d}(i_2)}{\mathrm{d}t} $$

If we integrate with respect to t (\$L_1\$ and \$L_2\$ are constants) we get

$$ L_1 i_1  =  L_2 i_2 $$

so the ratio of currents is

$$ \frac{i_1}{i_2} = \frac{L_2}{L_1} = \frac{2}{1} $$

In other word the biggest current will flow in the smallest inductor (just like the biggest voltage drop will be across the smallest capacitor in a series circuit). 

If you were to treat the inductance values as if they were "resistance values"  (which they are not) you would see immediately that "the bigger resistance" would take less current and it would be in the ratio of the "resistances"

To obtain the current \$i\$ combine the inductance values

$$ \frac{1}{L} = \frac{1}{L_1} + \frac{1}{L_2} $$

so that the circuit is reduced to a single inductor \$L = \frac{2}{3} H\$

then use kirchoff's law \$ i = i_1  + i_2\$

and as you already know \$ i1 = 2 \cdot i_2 \$  

you have everything you need to work it out.