(a) Calculate the initial and final kinetic energy (KE):

\(\text{KE (final)} = \frac{1}{2} \times 1500 \times 20^2 + \frac{1}{2} \times 750 \times 20^2\)

\(\text{KE (initial)} = \frac{1}{2} \times 1500 \times 30^2 + \frac{1}{2} \times 750 \times 30^2\)

Calculate the potential energy (PE) gain:

\(\text{PE gain} = 2250 \times 10 \times 800 \times 0.08\)

Calculate the work done against friction:

\(\text{WD against friction} = 600 \times 800\)

Use the energy equation:

\(\frac{1}{2} \times 2250 \times 30^2 + \text{DF} \times 800 = 600 \times 800 + \frac{1}{2} \times 2250 \times 20^2 + 2250 \times 10 \times 800 \times 0.08\)

Solve for DF:

\(\text{DF} = 1696.875 \text{ N} \approx 1700 \text{ N}\)

(b) Apply Newton's second law:

\(2400 - 600 = 2250a\)

Or, for the car and trailer separately:

\(T - 200 = 750a\) and \(2400 - 400 - T = 1500a\)

Solve for \(a\) and \(T\):

\(T = 800 \text{ N}\) and \(a = 0.8 \text{ m/s}^2\)