Since we are asked find the potential of the shell we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.
On the outside we can now:
- Distribute the internal charge $Q$ equally over the surface, without the external $Q$ present. That gives $V=Q/(4\pi\varepsilon_0|{\bf r}|)$.
- Subsequently put the external $Q$ at its position ${\bf r}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf r}_{\rm ext\_im}=(4R/3,0,0)$
- Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.
So the external potential is now known: $$ V_\text{ext}({\bf r}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf r}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} $$
On the inside we see nothing of what happens at the outside, we only have the charge $Q$ at position ${\bf r}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf r}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf r}) = \frac{Q}{4\pi\varepsilon_0|{\bf r}|-{\bf r}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{Q}{4\,\pi\varepsilon_0\,R} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.