(i) To find the speed of A at the instant B reaches the ground, we use Newton's Second Law and energy conservation.

Using Newton's Second Law for the system:

\(T - 0.35g = 0.35a\)

\(0.45g - T = 0.45a\)

Solving these equations gives:

\(0.45g - 0.35g = (0.35 + 0.45)a\)

\(0.1g = 0.8a\)

\(a = \frac{0.1g}{0.8} = 1.25 \text{ m/s}^2\)

Using the equation of motion:

\(v^2 = u^2 + 2as\)

where initial velocity \(u = 0\), \(a = 1.25 \text{ m/s}^2\), and \(s = 0.64 \text{ m}\):

\(v^2 = 2 \times 1.25 \times 0.64 = 1.6\)

\(v = \sqrt{1.6} = 1.26 \text{ m/s}\)

(ii) To find the total distance travelled by A after B reaches the ground, we consider the motion of A under gravity alone.

Using the equation of motion:

\(v^2 = u^2 + 2gs\)

where \(v = 0\), \(u = 1.26 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\):

\(0 = 1.26^2 - 2 \times 9.8 \times s\)

\(s = \frac{1.26^2}{2 \times 9.8} = 0.08 \text{ m}\)

The total distance travelled by A is twice this distance (as it moves up and then down):

\(\text{Total distance} = 2 \times 0.08 = 0.16 \text{ m}\)