(i) (a) Using Newton's Second Law for the system, we have:

\(1.3g - T = 1.3a\)

\(T - 0.7g = 0.7a\)

Adding these equations gives:

\(1.3g - 0.7g = (1.3 + 0.7)a\)

\(0.6g = 2a\)

\(a = 3 \text{ m/s}^2\)

Substituting back to find tension:

\(1.3g - T = 1.3 \times 3\)

\(T = 1.3g - 3.9\)

\(T = 9.1 \text{ N}\)

(b) Using the equation of motion:

\(s = \frac{1}{2} a t^2\)

\(2 = \frac{1}{2} \times 3 \times t^2\)

\(t^2 = \frac{4}{3}\)

\(t = \sqrt{\frac{4}{3}} \approx 1.15 \text{ seconds}\)

(ii) Using energy conservation or equations of motion:

At the point when the 1.3 kg particle reaches the plane, the speed \(v\) is given by:

\(v^2 = 2 \times 3 \times 2\)

\(v = \sqrt{12}\)

The 0.7 kg particle continues upwards:

\(0 = 12 - 2gs\)

\(s = \frac{12}{2g}\)

Greatest height above the plane is:

\(2 + \frac{12}{2g} = 4.6 \text{ m}\)