Why acceleration does not vary, when an equal force is applied in a pulley system? 
In the figure $m_{1} = 5$ Kg and $m_{2} = 2$ Kg and $F = 1$ N. Find the acceleration of either block. (The mass of the string and pulley is negligible and the pulley is frictionless)
  

I solved it as follows:
Let the tension in the string be $T$ and the acceleration be $a$. Then $$m_{1}g + F - T = m_{1}a$$ and
$$m_{2}g + F - T = -m_{2}a$$
Solving the two equations I got: $$a=\left( \dfrac {m_{1}-m_{2}}{m_{1}+m_{2}}\right) g$$ which is independent of $F$. Why the force $F$ gets cancelled out?
I was thinking that the force $F$ will cause different accelerations for the two blocks since $m_{1} \neq m_{2}$. If I would apply the force on block of mass $m_{1}$ then the acceleration for the system will be $a + a_{1}$ and then I apply the same force on the other block then the acceleration of the system will be $a +a_{1} - a_{2}$? which is certainly not equal to $a$ i.e. acceleration without the force $F$. 
 A: Think of it like a tug of war. If we are both pulling on each mass by the same amount, then those forces cancel out. 
This is why it's better to think about net forces rather than "net accelerations". What I mean by net acceleration is thinking about the acceleration each force would cause on that body and then combining them as if they were forces. It gets confusing here because you are right: each force would cause a different acceleration of each mass individually. 
But thinking about net forces and the system as a whole, we have $m_1g+F-T$ acting in one "direction", and $m_2g+F-T$ acting in the other "direction". Therefore, our net force is 
$$F_\text {net}=(m_1g+F-T)-(m_2g+F-T)=m_1g-m_2g$$
Notice how both $F$ and $T$ cancel out. This is because the forces cancel out in each direction. 
Keeping with thinking of the system as a whole we also have by N2L
$$F_\text {net}=(m_1+m_2)a$$
This leaves us with what you obtained.

It should be noted that this doesn't mean $F$ has no role here. It does affect the tension
$$T=F+\frac{2m_1m_2g}{m_1+m_2}$$
