(i) Applying Newton's second law to both particles:

For particle B:
\(0.45a = 0.45g - T\)

For particle A:
\(0.2a = T - F\)

Where \(F = 0.3 \times 0.2g\).

Combining the equations:
\((0.45 + 0.2)a = 0.45g - F\)

Substitute \(F\) and solve for \(a\):
\(a = 6\) m s^-2.

Using the equation \(v^2 = u^2 + 2as\) for particle B:
\(v^2 = 0 + 2 \times 6 \times (2 - (2.8 - 2.1))\)
\(v = 3.95\) m s^-1.

(ii) For particle A reaching the pulley:

Using \(v^2 = 3.95^2 + 2a_2[2.1 - (2 - 0.7)]\), where \(a_2 = -0.06g\).

Calculate \(v\):
\(v = 3.29\) m s^-1.

Alternative method for (ii):

Work done against friction:
\(WD = 0.06g \times [2.1 - (2 - 0.7)]\)

Using kinetic energy loss:
\(\frac{1}{2} \times 0.2 \times 3.95^2 - \frac{1}{2} \times 0.2 \times v^2 = 0.48\)
\(v = 3.29\) m s^-1.