(a)(i) The power developed by the engine is given by \(P = Fv\), where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force is \(440 + 280 = 720 \text{ N}\). Therefore, \(P = 720 \times 30 = 21600 \text{ W} = 21.6 \text{ kW}\).

(a)(ii) The power is decreased by 8 kW, so the new power is \(21600 - 8000 = 13600 \text{ W}\). The new driving force (DF) is \(\frac{13600}{30} = 453.33 \text{ N}\). Using Newton's second law for the system: \(DF - (440 + 280) = 2050a\). Solving for \(a\), we get \(a = \frac{453.33 - 720}{2050} = -0.13 \text{ m s}^{-2}\). For the tension \(T\) in the tow-bar, use \(T - 280 = 800a\), giving \(T = 176 \text{ N}\).

(b)(i) The driving force up the incline is \(DF = 720 + 2050 \times 0.06 = 1950 \text{ N}\). Using \(P = DF \times v\), we have \(28000 = 1950 \times v\), so \(v = \frac{28000}{1950} = 14.4 \text{ m s}^{-1}\).

(b)(ii) The increase in potential energy (PE) is \(PE = mgh\), where \(h\) is the height gained in 60 s. Using \(v = 14.4 \text{ m s}^{-1}\), the distance traveled is \(14.4 \times 60\). The height \(h = 14.4 \times 60 \times 0.06\). Therefore, \(PE = 800 \times 9.8 \times 14.4 \times 60 \times 0.06 = 414000 \text{ J} = 414 \text{ kJ}\).