The arrangement can be viewed as two spherical capacitors in series.

Assume the charges on the inner and outer surfaces of the middle spherical shell are $q_1$ and $q_2$. The total charge on the middle shell is $q_1+q_2=q$. The charges induced on the inner and outer shells are $-q_1$ and $-q_2$ respectively.

The PD between the inner and outer shell of a spherical capacitor is $kQ(\frac{1}{a}-\frac{1}{b})$ where $Q$ is the charge on the inner shell and $a,b$ are the radii of inner and outer shells. So in this case the PDs between inner and middle shells and between middle and outer shells are

$V_1=k(-q_1)(\frac{1}{r}-\frac{1}{2.5r})= -\frac35 \frac{kq_1}{r}$

$V_2=k(+q_2)(\frac{1}{2.5r}-\frac{1}{3.5r})=\frac{4}{35} \frac{kq_2}{r}$.

These PDs are equal (and opposite in sign) because the middle shell is common and the outer shells are at the same potential, so

$-V_1=V_2$

$\frac35 \frac{kq_1}{r}=\frac{4}{35} \frac{kq_2}{r}$

$q_2= \frac{21}{4}q_1$.

Substituting into $q_1+q_2=q$ gives

$q_1=\frac{4}{25}q$

$q_2=\frac{21}{25}q$.

The energies stored in capacitors 1 and 2 are

$U_1=\frac12q_1V_1=\frac{3}{10} \frac{kq_1^2}{r}$

$U_2=\frac12 q_2V_2=\frac{2}{35}\frac{kq_2^2}{r}$

Except for the discrepancy over the sign of $q_1$, which makes no difference to the answers, these expressions agree with the answers you have been given.

Isn't this a trick question? The electric charge on II resides on the outer surface. So no charge is induced on I. There is no electric field between conductors I & II, so no energy is stored in this region.

Do you have a worked solution?