The solution of Poisson's equation with appropriate boundary conditions is unique.
If you want spherical symmetry both the potential and the charge density must be functions of ##r## (in spherical coordinates only).
Thus 1) implies you have the solution
$$\phi(r)=\begin{cases} \frac{q}{4 \pi a} & \text{for} \quad r<a,\\
\frac{q}{4 \pi r} & \text{for} \quad r \geq a. \end{cases}$$
The charge distribution is
$$\rho=-\Delta \phi(r)=-\frac{1}{r} \partial_r^2 (r \phi) =-\frac{q}{4 \pi a r} \partial_r \Theta(a-r)=\frac{q}{4 \pi a^2} \delta(r-a). $$
This is a surface-charge distribution, ##\sigma=q/(4 \pi a^2)=\text{const}## along the sphere.
2) is somewhat more complicated. Let again ##a## be the radius of the inner conducting sphere and ##b## and ##c## the inner and outer radius of the shell of finite thickness, ##a<b<c##. For ##r<b## you have of course the same solution as above. Then you have
$$\phi(r)=\frac{q}{4 \pi} \begin{cases} 1/a & \text{for} \quad r < a, \\ 1/r & \text{for} \quad a \leq r<b, \\ 1/b & \text{for} \quad b \leq r<c, \\ 1/r & \text{for} \quad c \leq r. \end{cases}$$
You can calculate and interpret the various surface chrages on the surfaces of the conductors yourself!