(a) Since \((x-2)\) is a factor of \(p(x)\), substituting \(x = 2\) into \(p(x)\) gives:

\(8a - 40 + 2b + 8 = 0\)

\(8a + 2b - 32 = 0\)

\(4a + b = 16\) (Equation 1)

Since \((x-2)\) is also a factor of \(p'(x)\), differentiate \(p(x)\):

\(p'(x) = 3ax^2 - 20x + b\)

Substitute \(x = 2\) into \(p'(x)\):

\(12a - 40 + b = 0\)

\(12a + b = 40\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(12a + b - (4a + b) = 40 - 16\)

\(8a = 24\)

\(a = 3\)

Substitute \(a = 3\) into Equation 1:

\(4(3) + b = 16\)

\(12 + b = 16\)

\(b = 4\)

(b) With \(a = 3\) and \(b = 4\), \(p(x) = 3x^3 - 10x^2 + 4x + 8\).

Since \((x-2)\) is a factor, divide \(p(x)\) by \((x-2)\):

\(p(x) = (x-2)(3x^2 - 4x - 4)\)

Factorise \(3x^2 - 4x - 4\):

\(3x^2 - 4x - 4 = (3x + 2)(x - 2)\)

Thus, \(p(x) = (x-2)^2(3x + 2)\)

Final factorisation: \(3(x + \frac{2}{3})(x-2)^2\)