(N/A) According to Gauss's Law,the electric flux $\phi$ through a closed surface $S$ is given by $\phi = \oint_{S} \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_{0}}$.
$1$. If the total charge enclosed by a surface is zero $(q_{enclosed} = 0)$,it implies that the net electric flux through the surface is zero. However,this does not mean the electric field $\vec{E}$ is zero everywhere on the surface. The electric field at any point on the surface can be non-zero due to charges located outside the surface.
$2$. Conversely,if the electric field $\vec{E}$ is zero everywhere on the surface,then the surface integral $\oint_{S} \vec{E} \cdot d\vec{S}$ must be zero. According to Gauss's Law,this implies that the net charge enclosed by the surface must be zero $(q_{enclosed} = 0)$.