(a) The greatest speed is reached after accelerating for 5 seconds at 0.4 m/s². Using the formula:

\(v = u + at\)

where \(u = 0\), \(a = 0.4\), \(t = 5\):

\(v = 0 + 0.4 \times 5 = 2 \text{ m/s}\)

The velocity-time graph is a trapezium with a height of 2 m/s and time intervals of 5 s, 30 s, and 40 s.

(b) The total distance is the area under the velocity-time graph, which is a trapezium:

\(\text{Distance} = \frac{1}{2} \times (25 + 5 + 25) \times 2 = 65 \text{ m}\)

(c) Using Newton's second law during deceleration:

\(12250 - 1200g - mg = (1200 + m)(-0.2)\)

Solving for \(m\):

\(12250 = 1200 \times 9.8 + m \times 9.8 + 240 + 0.2m\)

\(12250 = 11760 + 240 + 9.8m + 0.2m\)

\(250 = 10m\)

\(m = 50\)

(d) During acceleration, using Newton's second law for the crate:

\(R - 50g = 50 \times 0.4\)

\(R = 50 \times 9.8 + 20\)

\(R = 520 \text{ N, upwards}\)