Ok so for the first part, its pretty simple. it just looks complicated than what it actually is (generally this is m1 all over). anyway, to resolve upwards...
Firstly, remember limiting friction is F = UR
ok so, it asks about P.
forget anything to do with the string. it is resting on the plane. so ignore the tension in the string.
resolving upwards,: R - 0.4gcos60 = 0
resolving across,: 0.4gsin60 - F = 0
it says in the question there is friction, and if P is more weight than Q, it's clearly gonna go down the plane when let go. so thats why friction on your diagram should be going UP the plane.
then to work out U its simple F / R by rearranging the 2 formulas i just gave you up there.
2. so if you add 0.2 to Q, its total mass is now 0.5 which is > than 0.4, so now the string will go the other way, and P will lift off the plane. So, we ignore the normal contact force and any other force acting on the plane, it just isn't there.
then resolving, remember just think logically where the particle is going to go, so for Q its 0.5g - T = 0.5a
for p its T - 0.4g = 0.4a
rearrange, to find A and T.
as soon as particles are in motion, F = MA needs to be used. as soon as you see acceleration anywhere, remember its F=MA