Laws of Motion

Q1) Take the reference frame that is fixed to the platform. In this reference frame there acts a pseudo force to the left on both the blocks: m₁a on the left block and m₂a on the right block.

If a < μg then obviously a < 2μg.
But then m₂a < μm₂g = f_2,max. Therefore, in this case the friction force f₂ between the mass m₂ and platform, which is a self adjusting force, acting on block 2 towards the right is enough to balance the (pseudo)force m₂a. And so it wont press against the block 1. As such the contact force between them will be zero. In this situation obviously the friction force on block 1 is also balancing m₁a.
As such, for the contact force between them to be non-zero which will happen when f₂ achieves its maximum value, we must have a ≥ μg.

However, lets see how does the blocks move if we do indeed increase a from its minimum value of a_min=μg.
Then, m₂ is pressing against m₁ though the system still does not move. The friction force between m₂ and the platform is μm₂g (since it has achieved its maximum value.) However, f₁ (friction between m₁ and platform) can still increase. Due to equilibrium of m₁ and m₂, we get
the contact force between them as
N = m₂(a-μg).
And
f₁=m₁a+N = (m₁+m₂)a - μm₂g
This force achieves its maximum when f₁ = 2μm₁g, i.e. when
a = m₂+2m₁m₁+m₂ μg

At this instant the mass m₁ gives way as well. How do u think the blocks will move after this?

Q2) The acceleration of each mass towards each other will be caused by the component of the tension in the thread. If the tension be T, then we have
2Tcosθ = F
giving T = F2cosθ
I have taken the angles shown as 30° as θ.
The component of this tension providing the acceleration towards each other is T sinθ. Therefore, the required acceleration
a = T sinθm = F tanθ2m = F2√3 m