(i) Since \((x + 2)\) is a factor of \(p(x)\), substituting \(x = -2\) into \(p(x)\) gives:

\(-16 + 4a - 2b - 4 = 0\)

\(4a - 2b = 20\) (Equation 1)

Differentiating \(p(x)\), we get \(p'(x) = 6x^2 + 2ax + b\).

Since \((x + 2)\) is a factor of \(p'(x)\), substituting \(x = -2\) into \(p'(x)\) gives:

\(24 - 4a + b = 0\)

\(-4a + b = -24\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 2: \(b = 4a - 24\)

Substitute into Equation 1:

\(4a - 2(4a - 24) = 20\)

\(4a - 8a + 48 = 20\)

\(-4a = -28\)

\(a = 7\)

Substitute \(a = 7\) into \(b = 4a - 24\):

\(b = 4(7) - 24 = 4\)

Thus, \(a = 7\) and \(b = 4\).

(ii) With \(a = 7\) and \(b = 4\), \(p(x) = 2x^3 + 7x^2 + 4x - 4\).

Since \((x + 2)\) is a factor, \((x + 2)^2\) is also a factor.

Attempt division by \((x + 2)^2\):

\(p(x) = (x + 2)^2(2x - 1)\)

Thus, the factorisation is \((x + 2)^2(2x - 1)\).