Solution

a) Heat dissipated in R when closing K

Initially: current I in $L_1$ zero in $L_2$
Magnetic fluxes cannot change abruptly. Upon closing K, the following is conserved:

$L_1 I_1 + L_2 I_2 = L_1 I$

In the final state,$I_1 = I_2 = I_0, so I_0 = \frac{L_1}{L_1+L_2}I$
The dissipated energy is the difference between the initial and final magnetic energy:

$W = \frac{1}{2}L_1 I^2 - \frac{1}{2}(L_1+L_2)I_0^2 = \frac{L_1 L_2}{2(L_1+L_2)} I^2$

b) Current limits after opening K

With K closed, currents are$I_1$in $L_1$ and$I_2$in $L_2$
Upon opening K, the total flux remains constant.
Since they are in series, $I_1' = I_2' = I'$
Therefore:

$(L_1 + L_2) I' = L_1 I_1 + L_2 I_2 \quad\Rightarrow\quad I' = \frac{L_1 I_1 + L_2 I_2}{L_1 + L_2}$

The current in L_1 varies between:

$I_1 \quad\text{and}\quad I_1 - \frac{2(I_1 - I_2)}{1 + L_1/L_2}$

The current in L_2 varies between:

$I_2 \quad\text{and}\quad I_2 + \frac{2(I_1 - I_2)}{1 + L_2/L_1}$