(i) The total distance fallen is calculated by finding the area under the velocity-time graph. The graph consists of a triangle, a trapezium, and a rectangle.

Area of the triangle (first stage):

\(\frac{1}{2} \times 5 \times 50 = 125 \text{ m}\)

Area of the trapezium (second stage):

\(\frac{1}{2} \times 7 \times (8 + 50) = 203 \text{ m}\)

Area of the rectangle (third stage):

\(90 \times 8 = 720 \text{ m}\)

\(Total distance = 125 + 203 + 720 = 1048 \text{ m}\)

(ii) During the second stage, the parachutist decelerates. The acceleration \(a\) is given by:

\(a = \frac{8 - 50}{12 - 5} = -6 \text{ m/s}^2\)

Using Newton's second law, \(F = ma\), where \(m\) is the mass of the parachutist:

\(850 - F = 85 \times (-6)\)

\(850 - F = -510\)

\(F = 850 + 510 = 1360 \text{ N}\)