This questions looks like it is defining position as the "x" component of the piston's position as the wheel spins it around. That x position is described by a simple harmonic motion expression. This is demonstrated by the gif at the top of this page:
As such, we can attempt to describe the position function with either of the standard equations of motion for simple harmonic oscillation in one dimension:
x(t) = Asin(ωt + φ) or Acos(ωt + φ)
A = amplitude of the motion (maximum displacement from equilibrium)
ω = angular frequency of the motion (in radians per second (rad/s))
φ = phase shift (this is generally not needed if, at t = 0, the oscillating object is at either the equilibrium position or one of the amplitudes)
If the wheel has radius A, then A is the amplitude, the farthest the object can get from equilibrium in the x direction (as shown in the gif). Since the object is at x = A at t = 0, we will not need the phase shift term φ.
What we do need is the form of the standard equation which gives x = A at t = 0. The sin version cannot do this, since t = 0 would give us sin(0), which is 0. But cos can, since at t = 0, we get cos(0) = 1. Thus, if:
x(t) = Acos(ωt), then x(0) = A, as we require.
So we will take x(t) = Acos(ωt)
From the problem:
A = 0.250 m
ω = 12 rad/s
(a) To answer this part, we just evaluate the above expression at t = 1.15 s.
x(1.15 s) = (0.250 m)cos(12 rad/s * 1.15 s)
[note: since the argument of the cos function is in radians, you need to have your calculator in "radians" mode to get the correct result]
(b) and (c) There is an associated sequence of linear velocity and acceleration functions which follow from x(t) = Acos(ωt). They are:
v(t) = -ωAsin(ωt)
a(t) = -ω2Acos(ωt)
(in calculus terms, these are just the first and second time derivatives of the position function)
To answer (b) and (c), just evaluate these expressions at t = 1.15 s. Remember to keep that calculator in radians! :)
I hope this helps. If you have any questions, please do not hesitate to ask. Best of luck!