(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) should yield zero. Thus, \(p\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right) + 2 = 0\).

Calculating, \(-\frac{1}{2}\) cubed is \(-\frac{1}{8}\), so \(4\left(-\frac{1}{8}\right) = -\frac{1}{2}\).

Thus, \(-\frac{1}{2} - \frac{a}{2} + 2 = 0\).

Solving for \(a\), we get \(-\frac{a}{2} = -\frac{3}{2}\), so \(a = 3\).

(ii)(a) With \(a = 3\), \(p(x) = 4x^3 + 3x + 2\). Performing polynomial division of \(4x^3 + 3x + 2\) by \(2x + 1\), we obtain \(2x^2 - x + 2\) as the quotient.

Thus, \(p(x) = (2x + 1)(2x^2 - x + 2)\).

(ii)(b) The critical value from \(2x + 1 = 0\) is \(x = -\frac{1}{2}\).

The quadratic \(2x^2 - x + 2\) is always positive since its discriminant \((-1)^2 - 4 \times 2 \times 2 = -15\) is negative, indicating no real roots and it opens upwards.

Thus, \(p(x) > 0\) for \(x > -\frac{1}{2}\).