(a) Surface charge density on the inner and outer shell. Taking a Gaussian surface of radius $r>r_1$ but $r < $ $r_2$. Since the Gaussian surface is inside the conductor, therefore electric field is zero everywhere

$$
\oint \text { E.dA }=\frac{q_{\text {enclosed }}}{\varepsilon_0}=\frac{q^{\prime}+\mathrm{q}}{\varepsilon_0}
$$
where $q^{\prime}$ be the charge on the inner shell surface.
$$
\begin{aligned}
&\Rightarrow \oint \mathrm{EdA}=0[\because \mathrm{E}=0] \\
&\Rightarrow \frac{\mathrm{q}^{\prime}+\mathrm{q}}{\varepsilon_0}=0 \quad \Rightarrow \mathrm{q}^{\prime}=-\mathrm{q}
\end{aligned}
$$
The conducting shell has no net charge, yet its inner shell has $-\mathrm{q}$ surface charge. Because the net charge on the shell is zero and no charge can be internal to the conductor, there must be $+q$ charge on the outer surface of the conductor, other than $+\mathrm{Q}$.
$\therefore$ Surface charge density of inner surface $=\frac{-\mathrm{q}}{4 \pi \mathrm{r}_1^2}$
And surface charge density of outer surface $=\frac{+\mathrm{Q}+\mathrm{q}}{4 \pi \mathrm{r}_2^2}$.
(b) By Gauss's law, the net charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. To dispose of this possibility, take a closed loop, part of which is inside the cavity along a field line and the rest
gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence there are no field lines inside the cavity i.e., no field, and no charge on the inner surface of the conductor, whatever be its shape.