A closed Gaussian surface contains no net charge, while some point...

According to Gauss's Law, the net electric flux through a closed surface is given by $\Phi = \oint \vec{E} \cdot d\vec{A} = \dfrac{q_{in}}{\epsilon_0}$.

Since the net charge enclosed by the Gaussian surface is zero ($q_{in} = 0$), the net electric flux $\Phi$ through the surface must be zero.

The electric field $\vec{E}$ at any point on the surface is the resultant field due to all charges present in space, both inside and outside the surface. Since there are point charges situated outside the surface, they will produce an electric field at points on the surface. Thus, the electric field $\vec{E}$ at the surface need not be zero, even though the net flux is zero.

Answer: The net electric flux $\Phi$ is zero, but $\vec{E}$ at the surface need not be zero.