A) The total energy of the system is sum of kinetic energy and elastic potential energy:

where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring
The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:

then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to

b) When the position of the object is

the potential energy of the system is

and so the kinetic energy is

since the mass is

, and the kinetic energy is given by

we can re-arrange the formula to find the speed of the object:

c) The potential energy when the object is at

is

Therefore the kinetic energy is

d) We already found the potential energy at point c, and it is given by