To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.
In this way,
The total mass of the cars would be,


Therefore the friction force at 29Km / h would be,




In this way the tension exerts between first car and locomotive is,



Therefore the tension in the coupling between the car and the locomotive is 
Answer:
Part a)

Part b)

Explanation:
Part a)
Since the diver is moving under gravity
so here its acceleration due to gravity will be uniform throughout the motion
so here we will have

here we have




Part b)
at highest point of his motion the final speed will be zero
so we will have



Answer:
9.21 m
Explanation:
because this is the answer