(2)

For a closed surface net electric flux ($\Phi$) depends upon the net charge enclosed. Mathematically $\Phi =\frac{q_i}{{\in }_0}$
 

(4)

Total flux from any closed surface is $\frac{1}{{\epsilon }_0}$ times net charge enclosed. So flux from any one surface of cube is $\frac{1}{6{\epsilon }_0}$ times total charge enclosed.

$=\frac{Q\mu C}{6{\epsilon }_0}=\frac{Q\times {10}^{-6}}{6{\epsilon }_0}$

(3)

The term $\oint \overrightarrow{E}·d\overrightarrow{s}$ represents electric flux over the closed surface and electric flux is proportional to the number of electric field lines. If $\oint \overrightarrow{E}·d\overrightarrow{s}=0$, it means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.