(i) The loss of potential energy (PE) of the system is calculated by considering the loss of PE of B and the gain of PE of A. The loss of PE of B is given by:

\(\text{Loss of PE of B} = 2g \times 3.24\)

The gain of PE of A is given by:

\(\text{Gain of PE of A} = 1.6g \times (3.24 \times 0.8)\)

Thus, the total loss of PE of the system is:

\(\text{Loss of PE} = 2g \times 3.24 - 1.6g \times (3.24 \times 0.8)\)

Substituting \(g = 9.8 \text{ m/s}^2\), we get:

\(\text{Loss of PE} = 2 \times 9.8 \times 3.24 - 1.6 \times 9.8 \times (3.24 \times 0.8) = 23.328 \text{ J}\)

(ii) Using the conservation of energy, the total kinetic energy (KE) gained by the system is equal to the total loss of potential energy:

\(\frac{1}{2} (1.6 + 2) v^2 = 23.328\)

Solving for \(v\):

\(1.8v^2 = 23.328\)

\(v^2 = \frac{23.328}{1.8}\)

\(v = \sqrt{12.96} = 3.6 \text{ m/s}\)