(i) The base of triangle \(XPQ\) is \(2 + p\) and the height is \(2p^2\) (since \(Q\) is on the curve \(y = 2x^2\) at \(x = p\), so \(y = 2p^2\)).

The area \(A\) of triangle \(XPQ\) is given by:

\(A = \frac{1}{2} \times (2 + p) \times 2p^2 = \frac{1}{2} \times (2p^2 + p^3)\)

\(A = p^3 + 2p^2\)

(ii) To find the rate of change of \(A\) with respect to time \(t\), we use the chain rule:

\(\frac{dA}{dt} = \frac{dA}{dp} \times \frac{dp}{dt}\)

First, find \(\frac{dA}{dp}\):

\(\frac{dA}{dp} = 3p^2 + 4p\)

Given \(\frac{dp}{dt} = 0.02\), substitute \(p = 2\):

\(\frac{dA}{dt} = (3(2)^2 + 4(2)) \times 0.02\)

\(\frac{dA}{dt} = (12 + 8) \times 0.02 = 20 \times 0.02 = 0.4\)