How do I analyse this pulley-trolley system? "find a force F such that the blocks are stationary relative to the trolley. All surfaces, wheels and pulley are assumed to be friction-less"
1: 
This is a very common mechanics problem and has been mentioned on this site before. I tried to solve this question from the pulley frame of reference using pseudo forces and it was quite easy to do so, but i cant wrap my mind around a few things.
1)Why is the tension dependent on the acceleration of the trolley?
the tension in string is found to be equal to $\frac  {m1m2(g+a)}{m1+m2}$ where a is the acceleration of the trolley relative to the ground. What is the intuition behind the it being dependent on a? Does this have something to do with the reaction force of pulley on the string
2) How can acceleration of the trolley prevent downward motion of block B?
if the trolley can only exert horizontal force on block B, how does it end up stopping it from falling for some value of a?
3) How do i effectively analyse this problem from the non inertial frame of reference without the use of pseudo forces?
 A: *

*Why is the tension dependent on the acceleration of the trolley?

The acceleration of the large block $M$ and the attached pulley does put a force on the string because the string must accelerate with the pulley. This increases the tension of the string and so creates a larger force on both blocks $m_1$ and $m_2$.


*How can acceleration of the trolley prevent downward motion of block B?

There are two ways to think of this. First, strings always exert tension along their length, so an increase in tension caused by the pushing of the pulley on the string creates an upward force on the hanging $m_2$ because that is the only direction the string can pull. Second, the string pulls on the pulley with equal magnitude down and to the left (tension is preserved across a massless, frictionless pulley). So, the reaction force must be up and to the right by Newton's Third Law. So, the pulley can create an upward force on the block $m_2$ which increases with acceleration.


*How do i effectively analyse this problem from the non inertial frame of reference without the use of pseudo forces?

The same way you would do any other physics problem:

*

*Draw a picture,

*Lay down a coordinate system,

*List each moving mass,

*List the forces on each mass,

*Write equations of motion for each mass using Newton's Second Law and any constraints,

*Solve the whole system.

Step 1 is done in the question.
For Step 2, let's use right and up as positive directions, left and down as negative directions.
Step 3, we have three masses: $M$, $m_1$, and $m_2$.
Step 4 and Step 5 is where the work starts.
The mass $M$ has three forces acting on it: the applied force $F$ to the right, the string tension $T$ to the left at the pulley, and the normal force to the left from block $m_2$.
$$F - T - F_N = Ma$$
where $a$ is the acceleration of $M$ (right is positive). Important note: the tension $T$ and the normal force $F_N$ have negative signs because they point to the left. This means that $T$ and $F_N$ should have positive values once you solve for them since they represent magnitudes.
Next, $m_1$ is acted on by the string tension $T$ to the right.
$$T = m_1a_1$$
where $a_1$ is the acceleration of $m_1$ (right is positive).
Finally, $m_2$ is acted upon by the string tension $T$ upward, gravity ($m_2g$) downward, and the normal force from block $M$ to the right.
$$T - m_2g = m_2a_2$$
$$F_N = m_2a$$
where $a_2$ is the acceleration of $m_2$ (up is positive) and $g$ is the acceleration due to gravity ($+9.8\,m/s^2$). We know that the tension on $m_1$ and $m_2$ is the same since the pulley is massless and frictionless (later, when you start studying rotational mechanics and torque, this will not necessarily be the case). We also know that the horizontal acceleration of the block $m_2$ is the same as the block $M$ because $m_2$ has no way of moving away ahead of block $M$.
We now have 5 unknowns ($a$, $a_1$, $a_2$, $F_N$, and $T$) and four equations. We have one more constraint equation due to $m_1$ and $m_2$ being tied together. This forces their accelerations relative to the block $M$ to be the same since their distance from each other must be constant.
$$a_2 = a - a_1$$.
If $a_1$ is larger than $a$, then the $m_2$ block will be accelerating downwards since the $m_1$ block will eventually overtake the $M$ block. Remember that $a_1$ is measured in the ground frame.
From here, you should be able to (Step 6) solve for the acceleration of $a_1$ in terms of the applied force $F$. Then, when you want to find when the blocks $m_1$ and $m_2$ are stationary with respect to the large block $M$, set $a_1 = a$ so that the block on top keeps up with the large block.
One more pedantic note, we can ignore the vertical forces on $M$ and $m_1$ and the horizontal forces on $m_2$ since these will be balanced by normal forces.
