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

Since \(p(x)\) is divisible by \((2x - 1)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) gives:

\(\frac{1}{4} + \frac{a}{4} - \frac{11}{2} + b = 0\)

Multiplying through by 4 to clear fractions:

\(1 + a - 22 + 4b = 0\)

\(a + 4b = 21\)

When \(p(x)\) is divided by \((x + 1)\), the remainder is 12. Substituting \(x = -1\) into \(p(x)\) gives:

\(-2 + a + 11 + b = 12\)

\(a + b = 3\)

Solving the system of equations:

1. \(a + 4b = 21\)

2. \(a + b = 3\)

Subtract equation 2 from equation 1:

\((a + 4b) - (a + b) = 21 - 3\)

\(3b = 18\)

\(b = 6\)

Substitute \(b = 6\) into \(a + b = 3\):

\(a + 6 = 3\)

\(a = -3\)

Thus, \(a = -3\) and \(b = 6\).

(i) Since \((x+1)\) and \((x+2)\) are factors of \(p(x)\), we have:

\(p(-1) = 0\) and \(p(-2) = 0\).

Substituting \(x = -1\) into \(p(x)\):

\(4(-1)^3 + a(-1)^2 + b(-1) - 2 = 0\)

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

\(a - b = 6\) (Equation 1)

Substituting \(x = -2\) into \(p(x)\):

\(4(-2)^3 + a(-2)^2 + b(-2) - 2 = 0\)

\(-32 + 4a - 2b - 2 = 0\)

\(4a - 2b = 34\) (Equation 2)

Solving Equation 1 and Equation 2 simultaneously:

From Equation 1: \(a = b + 6\)

Substitute into Equation 2:

\(4(b + 6) - 2b = 34\)

\(4b + 24 - 2b = 34\)

\(2b = 10\)

\(b = 5\)

Substitute \(b = 5\) into \(a = b + 6\):

\(a = 11\)

(ii) With \(a = 11\) and \(b = 5\), the polynomial is:

\(p(x) = 4x^3 + 11x^2 + 5x - 2\)

We need to find the remainder when \(p(x)\) is divided by \((x^2 + 1)\).

Using polynomial division or synthetic division, the quotient is \(4x + a\) and the remainder is \(x - 13\).

Since \((3x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{3}\) into \(p(x)\) should yield zero:

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

\(-\frac{1}{27}a + \frac{1}{9}b - \frac{1}{3} + 3 = 0\)

\(-\frac{1}{27}a + \frac{1}{9}b + \frac{8}{3} = 0\)

Multiply through by 27 to clear fractions:

\(-a + 3b + 72 = 0\)

\(a = 3b + 72\) (Equation 1)

When \(p(x)\) is divided by \((x - 2)\), the remainder is 21. Substitute \(x = 2\) into \(p(x)\):

\(p(2) = a(2)^3 + b(2)^2 + 2 + 3 = 21\)

\(8a + 4b + 5 = 21\)

\(8a + 4b = 16\)

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

Substitute \(a = 3b + 72\) from Equation 1 into Equation 2:

\(2(3b + 72) + b = 4\)

\(6b + 144 + b = 4\)

\(7b + 144 = 4\)

\(7b = -140\)

\(b = -20\)

Substitute \(b = -20\) back into Equation 1:

\(a = 3(-20) + 72\)

\(a = -60 + 72\)

\(a = 12\)

Thus, the values are \(a = 12\) and \(b = -20\).

(i) Since \((x + 2)\) is a factor of \(f(x)\), substituting \(x = -2\) into \(f(x)\) should yield zero:

\(f(-2) = (-2)^3 + a(-2)^2 - a(-2) + 14 = 0\)

\(-8 + 4a + 2a + 14 = 0\)

\(6a + 6 = 0\)

\(6a = -6\)

\(a = -1\)

(ii) With \(a = -1\), the polynomial becomes:

\(f(x) = x^3 - x^2 + x + 14\)

We need to show that \(f(x) = 0\) has only one real root. First, find the quadratic factor by dividing \(f(x)\) by \((x + 2)\):

\(f(x) = (x + 2)(x^2 - 3x + 7)\)

Now, use the discriminant of the quadratic \(x^2 - 3x + 7\):

\(b^2 - 4ac = (-3)^2 - 4(1)(7) = 9 - 28 = -19\)

Since the discriminant is negative, \(x^2 - 3x + 7 = 0\) has no real roots. Therefore, \(f(x) = 0\) has only one real root, which is \(x = -2\).