This problem can be solved by two methods : (A) using conservation of flux linkage and energy, and (B) using differential equations.
What you are missing in your solution is the conservation of energy. No energy is lost because there is no resistance in the circuit after the battery is disconnected.
(Note : With the type of switch depicted in the diagram, the circuit is broken - although only for a short time. The current through $L_1$ cannot then flow. This causes a large back emf and a spark across the switch, losing most of the energy stored in the inductor. To avoid loss of energy the switch needs to be smooth, with a small amount of overlap, so that there is never a break in the circuit containing the inductor.)
How the Circuit Behaves
While the switch is in position 1 there is a current $i_0=E/r$ flowing through $L_1$ and (presumably) none through $L_2$. It is possible that a sinusoidal current flows around circuit QRUT before the switch is thrown. However, we are not told anything about this so we must assume that there was initially no current in $L_2$.
When the switch is changed to position 2 the current $i_0$ continues to flow through $L_1$ - it cannot change instantaneously. For the same reason current cannot flow through $L_2$ immediately - it takes time to build up. Instead, all of the current from $L_1$ flows initially onto the capacitor.
There is initially no voltage across either inductor or the capacitor. As charge $q$ builds up on the capacitor there is an increasing voltage across all 3 components. This voltage increases the (upward) current $i_2$ through $L_2$ and decreases the (downward) current $i_1$ through $L_1$.
When the charge (and voltage) on the capacitor reaches a maximum, no more current flows into the capacitor. At this instant the current through $L_2$ is the same as that through $L_1$ - ie $i_1=i_2$ - because no current flows into or out of circuit PQTS. You can find the (maximum) charge on the capacitor at this instant using conservation of energy.
Although the charge and voltage on the capacitor then both decrease, this voltage continues to increase the (upward) current in $L_2$ and to decrease the (downward) current in $L_1$ until the charge on the capacitor becomes zero again. The current through $L_2$ is then a maximum because the polarity of the capacitor then switches, the voltage on it reverses, and this reduces the (upward) current $i_2$ while also increasing downward current $i_1$. By applying the conservation of energy again you can find the maximum current through $L_2$.
(A) Solution using Conservation of Energy
When the switch is thrown the current through $L_1$ is $i_0=E/r$. (This notation differs from yours.) Flux linkage is conserved in loop PQTS because this loop contains only inductors. When the currents in $L_1, L_2$ are generally $i_1, i_2$ we can write $$L_1i_0=L_1i_1+L_2i_2$$
When the charge on the capacitor is a maximum $q_0$ then $i_1=i_2$ so $(L_1+L_2)i_1=L_1i_0$. By conservation of energy we can write $$L_1i_0^2=L_1i_1^2+L_2i_2^2+\frac{1}{C}q^2=(L_1+L_2)i_1^2+\frac{1}{C}q_0^2=\frac{L_1^2}{L_1+L_2}i_0^2+\frac{1}{C}q_0^2$$ $$q_0^2=\frac{CL_1L_2}{L_1+L_2}i_0^2$$ As we shall find later we can write this in the form $$q_0=\frac{i_0}{\omega}$$ where $\omega$ is the angular frequency of oscillations in the circuit.
When the charge on the capacitor becomes zero again ($q=0$) then the current through $L_2$ reaches its maximum value $i_3$. Equations for the conservation of flux linkage and energy then become $$L_1i_0=L_1i_1+L_2i_3$$ $$L_1i_0^2=L_1i_1^2+L_2i_3^2=L_1i_0^2-2L_2i_0i_3+\frac{L_2^2}{L_1}i_3^2+L_2i_3^2$$ $$0=[(\frac{L_2}{L_1}+1)i_3-2i_0]L_2i_3$$ We have already met one solution $i_3=0$; the other solution is $$i_3=\frac{2L_1}{L_1+L_2}i_0$$ We can go on to find the minimum current through $L_1$ which is $$i_4=\frac{L_1-L_2}{L_1+L_2}i_0$$
(B) Solution using Differential Equations
The voltage across the components in the circuit are related to the currents and charges by $$V=-L_1\frac{di_1}{dt}=L_2\frac{di_2}{dt}=\frac{1}{C}q$$ $$i_1-i_2=\frac{dq}{dt}$$ From these we get the equations $$L_1i_1+L_2i_2=L_1i_0$$ $$\frac{d^2q}{dt^2}+\omega^2 q=0$$ $$\omega^2=\frac{1}{C}(\frac{1}{L_1}+\frac{1}{L_2})$$ with the solutions $$q=q_0\sin\omega t$$ $$(L_1+L_2)i_1=L_1i_0+L_2\frac{dq}{dt}=L_1i_0+L_2\omega q_0\cos\omega t$$ $$(L_1+L_2)i_2=L_1i_0-L_2\frac{dq}{dt}=L_1i_0-L_1\omega q_0\cos\omega t$$ Substitute $i_1=i_0$ when $t=0$ into the 1st eqn to get $$i_0=\omega q_0$$ We can then write general solutions for the currents as $$i_1=i_0\frac{L_1}{L_1+L_2}(1+\frac{L_2}{L_1}\cos\omega t)$$ $$i_2=i_0\frac{L_1}{L_1+L_2}(1-\cos\omega t)$$ The minimum current through $L_1$ and the maximum current through $L_2$ are respectively $$i_4=\frac{L_1-L_2}{L_1+L_2}i_0$$ $$i_3=\frac{2L_1}{L_1+L_2}i_0$$
Notes :
The current $i_2$ through $L_2$ always has a minimum of zero and is always positive (upward in the diagram).
Unlike $i_2$, the current $i_1$ through $L_1$ can become negative (flow upwards in the diagram). This happens if $L_1 \lt L_2$.
The charge on the capacitor is a maximum (both polarities) when the same current flows through both inductors - ie $i_1=i_2$.
When this happens the current through both inductors equals half the maximum current through $L_2$ - ie $$i_1=i_2=\frac{L_1}{L_1+L_2}i_0=\frac12 i_3$$
