(i) Applying Newton's second law to particle B, we have:

\(T - 1.6g = 1.6a\)

For particle A:

\(0.4g - T = 0.4a\)

Adding these equations gives:

\(1.2g = 2a\)

Thus, the acceleration \(a = 6 \text{ m/s}^2\).

Substituting back to find \(T\):

\(T = 1.6g - 1.6a = 1.6 \times 10 - 1.6 \times 6 = 6.4 \text{ N}\)

The work done by tension is:

\(\text{Work done} = T \times 1.2 = 6.4 \times 1.2 = 7.68 \text{ J}\)

(ii) Using the work-energy principle for particle A after B reaches the floor:

\(\frac{1}{2} \times 1.6 \times v^2 = 1.6 \times g \times 1.2 - 7.68\)

\(v^2 = 14.4\)

Using the equation of motion for A:

\(v^2 = 2gh\)

\(14.4 = 2 \times 10 \times h\)

\(h = 0.72 \text{ m}\)

The greatest height above the floor is:

\(H = 2 \times 1.2 + h = 2.4 + 0.72 = 3.12 \text{ m}\)