(i) Set the equations equal: \(x + 2 = 3\sqrt{x}\).

Rearrange: \(x - 3\sqrt{x} + 2 = 0\).

Let \(k = \sqrt{x}\), then \(k^2 - 3k + 2 = 0\).

Factorize: \((k-1)(k-2) = 0\).

So, \(k = 1\) or \(k = 2\).

Thus, \(\sqrt{x} = 1\) or \(\sqrt{x} = 2\).

Therefore, \(x = 1\) or \(x = 4\).

For \(x = 1\), \(y = 3\).

For \(x = 4\), \(y = 6\).

Coordinates of \(A\) are \((1, 3)\) and \(B\) are \((4, 6)\).

(ii) Find the area between the curves from \(x = 1\) to \(x = 4\).

Area = \(\int_1^4 (3\sqrt{x} - (x+2)) \, dx\).

Integrate: \(\int_1^4 (3x^{1/2} - x - 2) \, dx\).

\(= \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_1^4\).

Calculate: \(\left[ 16 - 8 - 8 \right] - \left[ 2 - \frac{1}{2} - 2 \right]\).

\(= 0 - (-1) = 1\).

Area = \(\frac{1}{2}\).