Since the object is moving at constant acceleration this is an uniformly accelerated motion and then we can use the following equations:

a)
In this case the initial velocity is 6.50 m/s, the final velocity is 13.0 m/s and the acceleration is 3.00 m/s², then we have:

Therefore, the displacement during this time interval is 21.23 m
b)
Since the object was moving in the positive direction from the beginning the distance is equal to the displacement; therefore, the distance during this time interval is 21.23 m
c)
In this case the initial velocity is -6.50 m/s, the final velocity is 13.0 m/s and the acceleration is 3.00 m/s², then we have:

Therefore, the displacement during this time interval is 21.23 m
d)
In this case we need to calculate the distance in two parts; first we need to calculate how much the object moves while its moving to the left until it stops; for this part the initial velocity is -6.50 m/s and the final velocity is zero, then we have:

Now we need to determine the displacement from when the velocity is zero until it reaches 13 m/s, then we have:

Finally we add the absolute value of this displacements to calculate the distance:

Therefore, the distance the object travelled is 35.21 m