All the points mentioned (a), (b) , (c) and (d) are correct.
let the train travel in positive x direction with a constant speed 10m/s. So the acceleration of the train is 0.
The speed of bouncing of the ball from the wall of the train compartment is the relative velocity of the ball wrt the speed of train. So at one collision the ball has a speed of 1 m/s in the positive x direction, wrt the train. So it will cover the length 10 m of the compartment in 10 sec. At each collision, the ball changes its direction to the opposite side. With respect to the observer from the ground too, the direction of the ball changes from - ve x axis to +ve x axis and vice versa.
speed of the ball is 10 + 1 = 11 m/s when the ball is travelling along positive x axis and 10 - 1 = 9 m/s when travelling in the -ve x axis direction. So the speed changes every 10 sec.
The average speed wrt stationary observer, of the ball during any 20 sec interval is
= [11 * 10 + 9 * 10 ] / 20 = 10 m/s
This remains same always. The motion of the ball is periodic, wrt an observer on the train. And, the average velocity is 0 and average speed = 1 m/s.
acceleration of the ball is same as that of the train, in case the train accelerates and does not travel at uniform speed. This is because the at each collision the wall will exert force on the ball to continue with the acceleration of the train.