(i) The power output is given by the formula:

\(P = F \times v\)

where \(P = 240\) W and \(v = 6\) m/s. Thus, the driving force \(F\) is:

\(F = \frac{240}{6} = 40\) N

Using Newton's Second Law:

\(F - R = ma\)

where \(m = 80\) kg and \(a = 0.3\) m/s². Therefore:

\(40 - R = 80 \times 0.3\)

\(R = 40 - 24 = 16\) N

(ii) For steady speed, the driving force equals the resistance:

\(\frac{240}{v} = 16\)

Solving for \(v\):

\(v = \frac{240}{16} = 15\) m/s

(iii) On an incline, using Newton's Second Law:

\(\frac{240}{4} - 16 - 80g \sin 3^{\circ} = 80a\)

\(60 - 16 - 80 \times 9.8 \times \sin 3^{\circ} = 80a\)

\(60 - 16 - 41.04 = 80a\)

\(2.96 = 80a\)

\(a = \frac{2.96}{80} = 0.0266\) m/s²