Why is surface charge distribution uniform for a conducting sphere? Can't it be arbitrary? 
If the charge $q_1$ has to repel the charge $q_2$, the electric field has to go inside the conductor which contradicts the fact that electric field inside conductors is zero. Then why do the charges distribute themselves in a particular manner? Why can't they be distributed over a small place on the conductor?  
 A: The electric field due to a charge surface element $dq$ does have to go inside the conductor, but it is cancelled out by other charges elsewhere on the sphere. So the field inside the conductor is zero, and there is no contradiction.
The precise nature of this cancellation can be seen by doing a surface integral, but it seems mysterious if you do it that way. If you apply Gauss's law, the precise cancellation becomes clear.
So the constant charge distribution satisfies $E=0$ inside the conductor. To show that it is unique, one needs the uniqueness theorems that are expounded upon in, for example, Griffiths Introduction to Electrodynamics.
A: Good question, but normally there are more than two charges.
Note that the electric field inside conductors is zero because the charges on the outside move to an arrangement where it is zero - or as close to zero as possible. 
Consider a sphere with a uniform density of charge on the outside. If the contributions of all the charges on the surface are summed up to calculate the electric field at any point inside the sphere the result is zero electric field.
A: The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that time, there is no electric field inside the conductor, and so no force on the charges that impels them to move to another, energetically more favorable, location.
A simple proof for spherical conductor is this: if the sphere is symmetrical, then the solution must also be symmetrical (there is nothing about a sphere that would drive an asymmetrical solution, and the uniqueness theorem says that if you have "a" solution that meets the boundary conditions, it must be "the" solution. Since uniform distribution meets the boundary conditions, it must be the solution.). But if that is so, then the electric field inside the sphere must also be spherically symmetrical. And we know from Gauss's theorem that the integral $\int E\cdot dS$ must equal the $\frac{Q}{\epsilon_0}$. Since $Q=0$, it follows that $E=0$.
