*M1 -- Magnitude of Force exerted on the PULLEY*

The force exerted on the pulley is the net vertical component of the forces in the string either side of the pulley.
So for (1):
F = 16.06sin30 + 16.06sin60
and for (2):
42.1sin30 + 42.1
..whatever they are equal to!

No mate those answers aren't correct unfortunately...the answers are:

(1) Force on pulley = T*(sqrt2) = 22.7N
which acts at an angle of 15 degrees to the vertical.

(2) Force on pulley = 2Tcos30 = 72.9N

Just to add that the questions are from:

(1) M1 Solomon Paper A
(2) M1 Solomon Paper H

for the force use the (botched to fit) cosine rule...

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)
[resultant force]^2 = 515.85 - 0
[resultant force] = 22.7 N

resultant force acts midway between angles from vertical because both forces are the same

(150 - 120)/2 = 15

reaction is always halfway, because thats the angle at which the pulley is in contact with the corner.

so look at attachment

pk

for the force use the (botched to fit) cosine rule...

[resultant force]^2 = [force1]^2 + [force2]^2 + 2[force1][force2]cos[angle between forces]

[resultant force]^2 = [16.06]^2 + [16.06]^2 + 2[16.06][16.06]cos(180-30-60)
[resultant force]^2 = 515.85 - 0
[resultant force] = 22.7 N