(i) Using the principle of conservation of energy, the kinetic energy (KE) gained by the particle is equal to the potential energy (PE) lost. The equation is:

\(\frac{1}{2} m v^2 - \frac{1}{2} m (7^2) = m g \times 5\)

Substituting the given values, \(m = 0.35\) kg, \(g = 10\) m/s²:

\(\frac{1}{2} \times 0.35 \times v^2 - \frac{1}{2} \times 0.35 \times 49 = 0.35 \times 10 \times 5\)

Solving for \(v\):

\(0.175 v^2 - 8.575 = 17.5\)

\(0.175 v^2 = 26.075\)

\(v^2 = 149\)

\(v = 12.2\) m/s

(ii) The work done against resistance is the difference between the initial mechanical energy and the final mechanical energy. The work done (WD) is given by:

\(\text{WD} = \text{PE loss} - \text{KE gain}\)

\(\text{WD} = m g h - \frac{1}{2} m (v_B^2 - v_A^2)\)

Substituting the values, \(h = 5\) m, \(v_A = 7\) m/s, \(v_B = 11\) m/s:

\(\text{WD} = 0.35 \times 10 \times 5 - \frac{1}{2} \times 0.35 \times (11^2 - 7^2)\)

\(\text{WD} = 17.5 - \frac{1}{2} \times 0.35 \times (121 - 49)\)

\(\text{WD} = 17.5 - 12.6\)

\(\text{WD} = 4.9\) J