Gauss's law in a uniform charge distribution extending infinitely in all directions Let us assume the universe filled with positive charge. About a particular point, all the positive charged particles will be symmetrical. Now consider a sphere of radius $r < \infty$ and apply Gauss law. According to Gauss law the electric flux through the sphere will be some finite value as it contains some positive charge. But by symmetry the electric field through the sphere and hence electric flux will be zero. This is a contradiction.
 A: The problem here is that you've failed to specify a boundary condition. 
In an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$, the answer is the solution to the set of differential equations 
$$\nabla \times \mathbf{E} = 0, \quad \nabla \cdot \mathbf{E} = \rho/\epsilon_0.$$
To get a unique solution to a differential equation, you have to specify a boundary condition. Usually, that condition is "the field is zero at infinity".
In this situation, the usual boundary condition doesn't work because the charge distribution is also infinite. To find the electric field, you must specify a boundary condition. Otherwise, the solution is just as ambiguous as trying to solve $F = ma$ without an initial position or velocity.
Once you do that, the symmetry will be broken, making your "$\mathbf{E}$ is symmetric so must be zero" argument fail. For example, one solution is $\mathbf{E}(\mathbf{r}) = kx \hat{i}$ for some value of $k$. You can check it satisfies Gauss's law. The boundary condition is "the field looks like $kx\hat{i}$ at infinity", which is not symmetric. 
This issue is subtle: it also comes up when considering gravitational fields in an infinite uniform universe. Newton made the same mistake, thinking that $\mathbf{g}$ had to vanish everywhere by symmetry.
A: The answer to this question completely depends upon how the charges are distributed.Saying that Gauss law holds here is correct but not in all cases. Suppose that the positive charges are to be randomly placed. Such a distribution would have slight non-uniformity, which would develop further with time because positive repels positive. So, a net flux will leave a sphere containing a point of symmetry.
Gauss law is falsified if all the charges are uniformly oriented instead, with no local non unifomity. In this case, all the charges would become the point of symmetry, hence no charge would remain special, hence no electric feild.
Let me modify this situation for simplicity. Let us assume a uniform positively charged medium spreading uniformly throughout the universe. Since we now assume the whole of universe to be uniformly charged, this is the same as no such charge anywhere in the universe at all.
Note that charging a uniform infinite medium uniformly is like not charging it at all! 
So, a universe with a uniform positive charge distribution is physically same as a universe with no such charge distribution at all. So Gauss law in the first universe appears to be violated, while in the second equivalent universe, it is not...
A: This is a good question because you are about to learn something. As outside charge create electric fields going in, it then also contributes fields going out on the opposite side of the surface. So they cancel themselves out. Where as the charge inside the surface only has electric fields going out. 
