Thank you for the opportunity to help you with your question!

Q1)Velocity from b-c is the slope of the line bc

We have the coordinate of b(1,2.2) and c(2,4.2)

Velocity=change in y/change in x =(4.2-2.2)/(2-1)=2m/s

From c-d

Coordinate of c(2,4.2) and d(3,2.2)

Velocity =(2.2-4.2)/(3-2)=-2m/s

From d-e

d(3,2.2) and e(4,2.8)

Velocity =(2.8-2.2)/(4-3)=.6 m/s

Q2)The speed first decreases and then increases from a to b.

Q3)Velocity at t=2 s is the y coordinate at x=2 which is 1m/s

Q4)From 3-5 s

from 3-4 s acceleration is the slope of the line =(4,2)and (3,0) =(2-0)/(4-3) m/s^2 =2 m/s^2

Acceleration from 4-5 seconds is 0 as the velocity is constant=2 m/s

Q5)Speed of the carmen=Speed of the train-Speed of the Juan as both are moving in the same direction so the relative speed can be found out by substracting both the speed. =20m/s+5m/s=25 m/s

Q6)Acceleration=sqrt(400^2/2*3500) =4.78 m/s^2

Q7)Distance=Speed*time =344*1.5=516 m

I hope this would help you,but if you have any doubt regarding this,you can surely ask.
:)