(i) Using Newton's second law, the net force on the particle while moving in the liquid is given by:

\(3g - R = 3 imes 5.5\)

Solving for the resistance \(R\):

\(R = 3g - 3 imes 5.5 = 3 imes 9.8 - 16.5 = 29.4 - 16.5 = 13.5 \text{ N}\)

(ii) To find the velocity at the surface, use the equation:

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

where \(u = 0\), \(a = 9.8 \text{ m/s}^2\), and \(s = 5 \text{ m}\):

\(v_s^2 = 2 imes 9.8 imes 5 = 98\)

\(v_s = \sqrt{98} = 10 \text{ m/s}\)

For the velocity at the bottom, use:

\(v_B^2 = v_T^2 + 2 imes 5.5 imes 4\)

where \(v_T = 10 \text{ m/s}\):

\(v_B^2 = 10^2 + 2 imes 5.5 imes 4 = 100 + 44 = 144\)

\(v_B = \sqrt{144} = 12 \text{ m/s}\)

For the time at the surface, use:

\(v = u + at\)

\(10 = 0 + 9.8t_1\)

\(t_1 = \frac{10}{9.8} \approx 1 \text{ s}\)

For the time at the bottom:

\(12 = 10 + 5.5(t_2 - t_1)\)

\(2 = 5.5(t_2 - 1)\)

\(t_2 - 1 = \frac{2}{5.5}\)

\(t_2 = 1 + \frac{2}{5.5} \approx 1.36 \text{ s}\)