Find the potential difference needed to reduce the kinetic energy of an electron by a factor of 5?

In part a) the kinetic energy is reduced from 100% to 0%. This corresponds to -57.6V (taking your answer as correct).

In part b) the kinetic energy is reduced from 100% to 20% (a factor of 5), which is an 80% loss of kinetic energy. The change is 80/100 = 4/5 of the energy change occurring in part a). So the answer is (4/5)x(-57.6)V.

In part c) kinetic energy is proportional to v². If the initial kinetic energy is 100%, the final kinetic energy is 100/5² = 4%, which is a 96% loss of kinetic energy. The change is 96/100 = 0.96 of the energy change occurring in part a). So the answer is 0.96x(-57.6)V.

I think you mean 4.5e6 m/s. This is very close to the speed of light, so the effects of relativity have to be taken into account. But let's ignore relativity first...

Find the kinetic energy of the particle:

K = (1/2)mv^2 (ignoring relativity)

K = 57.5eV

To stop it, a potential difference must be applied that will reduce the kinetic energy to 0. In other words, it will have to be negative.

V = -K/e = -57.5eV

This answer isn't correct though because relativity must be taken into account:

Find the relativistic kinetic energy:

K = sqrt( (pc)^2 + (mc^2)^2 ) - mc^2

p = mv = 4.19e-24 kg m/s

K = 57.6eV

V = -K/e = -57.6V