Simple Harmonic motion

1)
Since the spring is unstretched when the collar is at A, this means AC is the natural length of the spring, 0.5m.
You first need to find the length of BC in order to determine the initial force acting on the collar at B. The angle BAC is 120° which I hope you know.

By cosine laws,
BC² = AB² + AC² - 2(AB)(AC)cos(120°)
BC² = 0.5² + 0.5² - 2(0.5)(0.5)(-1/2)
BC = 0.866m

Therefore, the force acting on the collar by the spring at B along the rod direction is F=kx=150(0.866-0.5)cos(30°)=47.55N
The force acting on the collar by gravitational force along the rod direction is
F=mg=4(10)cos(30°)=34.64N

These two forces are in opposite direction.
Hence, the acceleration is a=F/m=(47.55-34.64)/4 = 3.23m/s²

2)a)
This question can be solved by conservation of energy.
As the springs are not distorted when the mass is at A, it implies that A is the equilibrium position.

By conservation of energy, all potential energy of the springs converts to kinetic energy at equilibrium point,
8x0.2²/2 + 5x0.2²/2 = 4v²/2
v=0.361m/s

2)b)
The initial position of the mass (displaced at 20cm at B) has fixed the amplitude of the oscillation, so the maximum displacement the mass can reach on the left hand side is also 20cm.