(i) The work done by a force is given by \(W = Fd \cos \theta\), where \(F\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle between the force and the direction of movement.

The work done by the 30 N force is \(30 \times 20 \times 0.6\).

The work done by the 40 N force is \(40 \times 20 \times 0.8\).

Total work done = \(30 \times 20 \times 0.6 + 40 \times 20 \times 0.8 = 1000 \text{ J}\).

(ii) Since the block is moving with constant speed, the net force in the x-direction is zero. The frictional force \(F_f = \mu W\), where \(\mu = \frac{5}{8}\) and \(W\) is the weight of the block.

The equation for equilibrium in the x-direction is:

\(30 \times 0.6 + 40 \times 0.8 - \frac{5}{8}W = 0\).

Solving for \(W\):

\(18 + 32 - \frac{5}{8}W = 0\)

\(50 = \frac{5}{8}W\)

\(W = \frac{50 \times 8}{5} = 80 \text{ N}\).