Solution:

Total gravitational potential at a point \( r = \frac{a}{2} \):

1. Potential due to the mass at the center:
The gravitational potential at a distance \( r \) from a point mass \( M \) is given by:
\[
V_{\text{center}} = -\frac{GM}{r}
\]
At \( r = \frac{a}{2} \), this becomes:
\[
V_{\text{center}} = -\frac{GM}{\frac{a}{2}} = -\frac{2GM}{a}
\]

2. **Potential due to the spherical shell**:
Inside a spherical shell, the gravitational potential is constant and equal to the potential at the surface, which is:
\[
V_{\text{shell}} = -\frac{GM}{a}
\]

### Total potential at \( r = \frac{a}{2} \):

The total gravitational potential is the sum of the potentials due to the mass at the center and the shell:
\[
V_{\text{total}} = V_{\text{center}} + V_{\text{shell}} = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the gravitational potential at a distance \( \frac{a}{2} \) from the center is:
\[
V = -\frac{3GM}{a}
\]