If a source charge exerts force on the test charge, it should also feel the same force in opposite direction. So, won't it mean source charge is experiencing force due to its own electric field?
There will be force on the source charge, but due to electric field of the test charge. If the test charge was not present, there would be no force on the source charge.
For an INFINITE parallel plate capacitor, the electric field has the same value everywhere between the 2 plates. An intuitive reason for that is: suppose you have a small test charge +q at a distance $x$ away from the +ve plate and a distance $d-x$ away from the -ve plate. The +ve plate will repel the charge and the -ve plate will attract it. Now if the charge is at a distance $y$ from the +ve plate $(y>x)$ then the the repulsive force due to the +ve plate will be weaker but the attractive force due to the -ve plate will be stronger such that the net force on the test charge will have the same value as when the charge was a distance $x$ away from the +ve plate. This is because the electric field lines due to an infinite plate are parallel to the normal vector to the plate.
I don't quite understand your second question. If you have a fixed potential difference between the two plates, then the electric field between the plates is given by $E=\frac{V}{d}$, where d is the distance between the 2 plates. From the expression, you can see that the electric field is NOT independent of the separation of the 2 plates.
Now suppose you keep the charge on both plates the same and separate them, such the the distance, $d$ increases. In that case, the electric field will remain unchanged. This might seem counter intuitive at first sight, this might seem like a violation of the conservation of energy. But, the extra energy required to keep the E field the same comes from the work that you are putting in to separate the attracting plates. The capacitance, $C$=$\frac{\epsilon A}{d}$, where A is the area of the plates, d is the separation of the plates and $\epsilon$ is the permittivity of the material between the plates. If $\epsilon$ and $A$ are constants, then, $C$ is inversely proportional to $d$.
The relation between the potential difference between the 2 plates, $V$ and the charge on one of the plates, $Q$ is given by $Q=CV=\frac{\epsilon A V}{d}=\epsilon A E$.
$\implies E=\frac{Q}{\epsilon A}$
Therefore, E is independent of $d$ if the charge on each plate is unchanged.
Best Answer
The electric field is a vector, a quantity that has both a magnitude and a direction. The electric field intensity is the magnitude of the vector.
For example, if we had an electric field vector which extended 1 unit in the x direction and 1 unit in the y direction, then its magnitude would be $\sqrt{1^2+1^2}=\sqrt{2}$ units.