3: At first contact, the velocity downward will be v=sqrt(2gx) (see formula 2.4.4c in Lec02), or 4.43 m/s. The downward motion then stops in 0.02 s, giving an average acceleration of 221 m/s2. Now F=ma, so the force is 70x221=15500 N. (The answer should include the weight of the man, +70x9.8=686 N).
4: The whole train moves with the same acceleration. In order for the last 8 boxcars to have that acceleration, the NET force on car 38 must be F = ma = 8x55000x0.05 N. The coupling must pull with this force PLUS the 8x16000 N friction. The coupling has an equal and opposite force on car 37.