(i) The work done by the man is calculated using the formula for work done by a force at an angle:

\(WD = F \cdot d \cdot \cos \theta\)

where \(F = 35 \text{ N}\), \(d = 12 \text{ m}\), and \(\theta = 20^\circ\).

Thus,

\(WD = 35 \cdot 12 \cdot \cos 20^\circ = 395 \text{ J}\)

(ii) To find the speed attained by the wheelbarrow, we can use the work-energy principle or Newton's second law.

Using the work-energy principle:

The work done against resistance is \(15 \times 12 = 180 \text{ J}\).

The net work done is used to increase the kinetic energy:

\(35 \cdot 12 \cdot \cos 20^\circ - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

Solving for \(v\):

\(395 - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

\(215 = 12.5 \cdot v^2\)

\(v^2 = \frac{215}{12.5}\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)

Alternatively, using Newton's second law:

The net force is \(35 \cos 20^\circ - 15 = 25a\).

Solving for \(a\):

\(a = \frac{35 \cos 20^\circ - 15}{25} \approx 0.716 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\):

\(v^2 = 2 \cdot 0.716 \cdot 12\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)