1.

We know that PV = 1/3 Nmc^2 and PV = NkT (where c is the mean velocity of gas particles)

We can simplify this to:

mc^2 = 3kT

And

1/2 mc^2 = 3/2 kT

Which means that

K.E. is directly proportional to temperature T.

2.

Resolve the tensions in each string.The vertical components of both tensions add up to Fg. The horizontal components of both tensions add on to zero (each is equal to a certain F that I introduced).

The vertical component provided by one string is Tsin(45) = Fg/2.

The same vertical component is provided by the other string.

Along the horizontal, we have Tcos(45) = F

And T^2 = (Fg/2)^2 + F^2

Of course, that's the 'calculation' part of it. If you can 'see' it right from the start, you get the answer easily. But that's something you see after quite some practice.

3.

Find the potential energy of the stone.

Hence, find the kinetic energy the stone normally gets after falling 3 m without air resistance.

Find the kinetic energy that the stone actually has, after accounting for air resistance.

Convert kinetic energy to a speed.

Can you give those a try?

[Note that there is nothing involving torque whatsoever in those questions.]