(i) Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(8\left(-\frac{1}{2}\right)^3 + a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + 3 = 0\)

Simplifying, we get:

\(-1 + \frac{a}{4} - \frac{b}{2} + 3 = 0\)

\(a - 2b + 8 = 0\)

When \(p(x)\) is divided by \((2x - 1)\), substituting \(x = \frac{1}{2}\) gives a remainder of 1:

\(8\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 3 = 1\)

Simplifying, we get:

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

\(a + 2b + 12 = 0\)

Solving the equations \(a - 2b + 8 = 0\) and \(a + 2b + 12 = 0\), we find:

\(a = -10\), \(b = -1\)

(ii) With \(a = -10\) and \(b = -1\), divide \(p(x) = 8x^3 - 10x^2 - x + 3\) by \(2x^2 - 1\):

The division gives a quotient of \(4x - 5\) and a remainder of \(3x - 2\).