(i) Since \((x + 1)\) is a factor of \(p(x)\), substituting \(x = -1\) into \(p(x)\) gives:

\(-8 + a - b - 1 = 0\)

\(a - b = 9\) (Equation 1)

When \(p(x)\) is divided by \((2x + 1)\), the remainder is 1. Substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1\)

\(\frac{1}{4}a - \frac{1}{2}b = 3\) (Equation 2)

Solving Equation 1 and Equation 2 simultaneously:

From Equation 1: \(a = b + 9\)

Substitute into Equation 2:

\(\frac{1}{4}(b + 9) - \frac{1}{2}b = 3\)

\(\frac{1}{4}b + \frac{9}{4} - \frac{1}{2}b = 3\)

\(-\frac{1}{4}b = 3 - \frac{9}{4}\)

\(-\frac{1}{4}b = \frac{3}{4}\)

\(b = -3\)

Substitute \(b = -3\) into \(a = b + 9\):

\(a = 6\)

(ii) With \(a = 6\) and \(b = -3\), the polynomial is:

\(p(x) = 8x^3 + 6x^2 - 3x - 1\)

Since \((x + 1)\) is a factor, divide \(p(x)\) by \((x + 1)\):

\(8x^3 + 6x^2 - 3x - 1 = (x + 1)(8x^2 - 2x - 1)\)

Factor the quadratic \(8x^2 - 2x - 1\):

\(8x^2 - 2x - 1 = (4x + 1)(2x - 1)\)

Thus, \(p(x) = (x + 1)(4x + 1)(2x - 1)\).