Hooke's law says that F = -kx, where
x = distance the spring has been stretched or compressed from equilibrium
F = is the restoring force exerted by the spring
k = the spring constant (with units of force/length)
So if the spring was compressed to 7.00 cm instead of 3.5 cm, it will have the maximum F_2 will be 2 times as great as before, i.e. F_2 = 2F. Also, the average force will be 2 times as great (the force decreases linearly over the distance as it returns to the equilibrium position). But the kinetic energy of the wood would be the same as the potential energy when the spring is at it's maximum compression. Pe_2 = F_2*d_2 = F_2*x_2. Since this F_2 = 2F and x_2 = 2x, this gives PE_2 = 4*F*x. So the answer is K_2 = 4 * K.
2)Or you could do it more directly by knowing that the potential energy of a spring is given by the formula
PE = (1/2)kx^2
Since x_2 = 5x,
PE_2 = (1/2)kx_2^2
PE_2 = (1/2)k(2x)^2
PE_2 = 4*(1/2)kx^2
PE_2 = 4*PE
Since the potential energy is 4 times as great, and since it is all converted to kinetic energy, K_2 = 4 K.