(i) To find the height of A above the surface of the liquid, use the area under the velocity-time graph from t = 0 to t = 0.7 s. The area represents the displacement:

Area = \(\frac{1}{2} \times 0.7 \times 7 = 2.45\) m.

Thus, the height of A above the surface of the liquid is 2.45 m.

\(To find the depth of the liquid, use the area under the graph from t = 0.7 to t = 1.2 s:\)

Area = \(\frac{1}{2} \times (7 + 5) \times (1.2 - 0.7) = 3\) m.

Thus, the depth of the liquid is 3 m.

(ii) The deceleration of the stone while it is moving in the liquid can be found using the gradient of the velocity-time graph from t = 0.7 to t = 1.2 s:

Deceleration = \(\frac{5 - 7}{1.2 - 0.7} = -4\) m s^-2.

(iii) Using Newton's second law, the net force on the stone is given by:

\(0.7 - mg = ma\)

where \(a = -4\) m s^-2 and \(g = 9.8\) m s^-2.

Rearranging gives:

\(0.7 = m(9.8 - 4)\)

\(m = \frac{0.7}{5.8} = 0.05\) kg.