= Solution
When $a\ne0$ and $n\geq2$, the set of achievable mean-variance pairs is
$$
\boxed{A=\{(m,s^2):m\in\mathbb R,\ s^2\geq(m-b)^2/D\}.}
$$
Indeed 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 $(b,s^2)$, $s^2\geq0$.
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 $a\ne0$ it is \b[the upper branch $m\geq b$, $s^2=(m-b)^2/D$]. 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.
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