We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

Using (2) we can rewrite this as

And using (1), we find

Substituting
into the last equation, we find the value of x:

B) 
In this case, the kinetic energy is 1/10 of the total energy:

Since we have

we can write

And so we find:

Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N