(i) For block B in equilibrium, the tension in S₂ must balance the weight of both blocks. Therefore, the tension in S₂ is given by:

\(T_{S_2} = (3 + 2)g = 5g = 50 \text{ N}\)

For block A, the tension in S₁ must balance the weight of block A alone. Therefore, the tension in S₁ is:

\(T_{S_1} = 3g = 30 \text{ N}\)

(ii) After S₂ breaks, apply Newton's second law to each block. For block A:

\(3g - T - 1.6 = 3a\)

For block B:

\(2g + T - 4 = 2a\)

Adding these equations:

\((3g + 2g) - (1.6 + 4) = (3 + 2)a\)

\(5g - 5.6 = 5a\)

\(a = \frac{5g - 5.6}{5} = 8.88 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations to find \(T\):

\(3g - T - 1.6 = 3 \times 8.88\)

\(T = 3g - 1.6 - 26.64\)

\(T = 1.76 \text{ N}\)