X(t) = t^3 - 12t^2 + 36t velocity = x'(t) = 3y^3 - 24t + 36 = 0 x'(t) = t^2 - 8t + 12 = 0 x'(t) = t^2 - 6t - 2t + 12 = 0 x'(t) = t(t - 6) - 2(t - 6) = 0 x'(t) = (t - 2)(t - 6) = 0 t = 2 and t = 6 must be the roots. is my answer of d "2 and 6" correct based on this work?

Yes you have the right idea. You find the derivative to get the velocity function and then determine when the velocity is zero. That turns out to be at t = 2 and t = 6. Both are correct. The final answer is correct.

The only flaw I see is that you wrote 
x'(t) = 3y^3 - 24t + 36
on line 2 when it should be
x'(t) = 3t^2 - 24t + 36

Other than that it looks perfect.

What’s the improper fraction?

Yes you do, so £1973.09 - £1473.50 = £499.59 so tim overpaid £499.59 .hope this ; )