(i) Since \((x - 2)\) is a factor of \(p(x)\), substituting \(x = 2\) into \(p(x)\) should yield zero: \(2(2)^3 + a(2)^2 - 4 = 0\). Simplifying gives \(16 + 4a - 4 = 0\), leading to \(4a = -12\), so \(a = -3\).

(ii) With \(a = -3\), \(p(x) = 2x^3 - 3x^2 - 4\). Dividing by \((x - 2)\) gives the quadratic factor \(2x^2 + x + 2\). Thus, \(p(x) = (x-2)(2x^2 + x + 2)\).

(iii) To solve \(p(x) > 0\), note that \(2x^2 + x + 2\) has no real roots and is always positive. Therefore, \(p(x) > 0\) when \(x > 2\).