A spring with a spring constant k is connected to the ceiling. First a weight of mass m is connected to the spring. Deduce the new equilibrium position of the spring, find its equation of motion and hence deduce its frequency f.

Let's first find the equilibrium position of the spring. When the mass is first attached the spring will oscillate like a simple harmonic oscillator, in the real world the oscillator will eventually settle and this position will be the new equilibrium position of the oscillator. Let's make the velocity 0 and resolve Newton's Second Law vertically (N2L from here on in): For a spring F=kx, N2L vertically kx=mg. Yielding x=mg/k where g is the local acceleration due to gravity.
For the Equation of Motion, resolve N2L while the oscillator is in motion. To make life easier lets substitute x = x0+y where x0 is the new equilibrium position as this way we don't need to worry about any constants and can have y oscillate around y=0. Now: ma=-kx where a = d^2x/dt^2. Solving this differential equation we get y = Acos(wt) + Bsin(wt) where we find that w^2 = k/m. Knowing that w=2pif we can work out the frequency of these oscillations.