As depicted in the figure, a charge q is situated at the centre of...

The initial electric flux through the curved surface of the hemisphere due to the charge $q$ at the centre is $\Phi_{q} = \dfrac{q}{2\epsilon_0}$.

When a second charge $Q$ is placed at any external point, the total flux through the hemisphere becomes $\Phi_{\text{total}} = \Phi_{q} + \Phi_{Q}$, where $\Phi_{Q}$ is the flux due to the charge $Q$.

For the total flux to remain unchanged, the flux through the curved surface due to $Q$ must be zero, i.e., $\Phi_{Q} = 0$.

Consider a closed surface formed by the curved hemispherical surface and its flat circular base. By Gauss's law, the net flux through this closed surface due to an external charge $Q$ is zero.

$\Phi_{\text{closed}} = \Phi_{\text{curved}} + \Phi_{\text{base}} = 0 \Rightarrow \Phi_{\text{curved}} = -\Phi_{\text{base}}$

For points $A$ and $C$, the charge $Q$ lies exactly on the plane containing the flat base of the hemisphere. The electric field lines produced by $Q$ at any point on the base will lie in the plane of the base.

Since the area vector of the base is perpendicular to its plane, the electric field $\vec{E}$ is perpendicular to the area vector $d\vec{A}$ everywhere on the base.

Thus, the flux through the base is $\Phi_{\text{base}} = \int \vec{E} \cdot d\vec{A} = 0$.

Consequently, $\Phi_{\text{curved}} = 0$ for charges placed at $A$ and $C$.

For points $B$ and $D$, the charge is not in the plane of the base, so field lines will cross the base, making $\Phi_{\text{base}} \neq 0$ and thus $\Phi_{\text{curved}} \neq 0$.

Therefore, the flux through the hemisphere remains unchanged only when the second charge is placed at locations $A$ and $C$.

Answer: $A$; $C$