 Solution

When $a\ne 0$ and $n\ge 2$, the set of achievable mean-variance pairs isIndeed the decomposition in part (a) proves necessity. There is a nonzero $h$ orthogonal to $a$; scaling it makes $h^{\mathsf{T}}Vh$ any nonnegative number, proving sufficiency. For $n=1$ that orthogonal space is zero and only the boundary parabola is achievable, not its interior. If $a=0$, the set is the vertical half-line $(b,s^2)$, $s^2\ge 0$.

The mean-variance efficient frontier comprises undominated portfolios: no other portfolio has at least as high a mean and at most as large a variance, with one strict inequality. For $a\ne 0$ it is the upper branch $m\ge b$, $s^2=(m-b{)}^2/D$. Lower-branch portfolios have lower mean and larger variance than the all-bank portfolio, and points above the minimum-variance boundary are dominated at their own mean. On the upper branch increasing mean necessarily increases minimum variance, so no dominance is possible. If $a=0$, only the zero-variance all-bank point is efficient.