What kind of Gaussian surface is used to calculate the electric field due to the charge distribution?

$(a)$ An electrostatic field line is a continuous curve. That is,a field line cannot have sudden breaks. Why not?
$(b)$ Explain why two field lines never cross each other at any point?

$A$ cubical Gaussian surface has a side of length $a = 10 \,cm$. Electric field lines are parallel to the $X$-axis as shown in the figure. The magnitudes of the electric fields through surfaces $ABCD$ and $EFGH$ are $6 \,kNC^{-1}$ and $9 \,kNC^{-1}$ respectively. Then,the total charge enclosed by the cube is (Take $\varepsilon_0 = 9 \times 10^{-12} \,Fm^{-1}$): (in $\,nC$)

The adjoining diagram shows the electric lines of force emerging from a charged body. If the electric fields at $A$ and $B$ are $E_A$ and $E_B$ respectively and the distance between them is $r$,then

If some charge is given to a solid metallic sphere,the field inside remains zero and by Gauss's law all the charge resides on the surface. Now,suppose that Coulomb's force between two charges varies as $1 / r^{3}$. Then,for a charged solid metallic sphere

Consider a cube of side length '$a$' centered at the origin. It is enclosed by three fixed point charges: $(-q)$ at $(0, -a/4, 0)$,$(+3q)$ at $(0, 0, 0)$,and $(-q)$ at $(0, a/4, 0)$. Choose the correct option.