(a) The potential energy lost in 50 m is given by:

\((m + 300) g \times 50 \sin 3^{\circ}\)

Using the work-energy principle, we have:

\((m + 300) g \times 50 \sin 3^{\circ} - 40000 = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

Alternative method:

The resistance force \(R\) is:

\(R = \frac{40000}{50} = 800\) N

Using Newton's second law:

\((m + 300) g \sin 3^{\circ} - R = 0\)

Solving for \(m\):

\(m = 1230\) kg to 3 significant figures.

(b) Applying Newton's second law to the trailer:

\(T + 300 g \sin 3^{\circ} - 200 = 0\)

Or to the car:

\(mg \sin 3^{\circ} = T + 600\)

Solving for \(T\):

\(T = 43.0\) N to 3 significant figures.