A 8000kg engine pulls a train …

Force exerted by the engine,F' = 40,000 N
Frictional force  offered by the track in the direction opposite of the motion,F'' = -5,000 N
(a) The net accelerating force, F = F'+F''= 40,000 N + (-5,000 N) = 35,000 N
(b) Mass of each wagon of the train = 2000 kg

Number of wagons = 5

Therefore, Mass of the train,m= 2,000 kg  $\times$ 5= 10,000 kg.

Net accelerating force acting on the train,F= 35000 N 
From Newton's second law of motion, acceleration
$a = \frac{F}{m} = \frac{{35,000\,N}}{{10,000\,kg}} = 3.5\,\,m{s^{ - 2}}$
(c)Mass of 1 wagon= 2000 kg

acceleration of the train= 3.5 ms^-2

From the relation,F =ma,we get

F= 2000 kg $\times$ 3.5 ms^-2

F= 7000 N

Force exerted by wagon 1 on wagon 2

=Net accelerating force -  Force acting on wagon 1= 35000 N - 7000 N= 28000 N

Therefore,the required answer is 28000 N