(i) Evaluate the function \(f(x) = x^3 - 8x - 13\) at \(x = 3\) and \(x = 4\):

\(f(3) = 3^3 - 8 \times 3 - 13 = 27 - 24 - 13 = -10\)

\(f(4) = 4^3 - 8 \times 4 - 13 = 64 - 32 - 13 = 19\)

Since \(f(3) < 0\) and \(f(4) > 0\), the root lies between 3 and 4.

(ii) Use the iterative formula \(x_{n+1} = (8x_n + 13)^{\frac{1}{3}}\):

Start with an initial guess, \(x_0 = 3.5\).

\(x_1 = (8 \times 3.5 + 13)^{\frac{1}{3}} = (41)^{\frac{1}{3}} \approx 3.4482\)

\(x_2 = (8 \times 3.4482 + 13)^{\frac{1}{3}} = (40.5856)^{\frac{1}{3}} \approx 3.4325\)

\(x_3 = (8 \times 3.4325 + 13)^{\frac{1}{3}} = (40.46)^{\frac{1}{3}} \approx 3.4301\)

\(x_4 = (8 \times 3.4301 + 13)^{\frac{1}{3}} = (40.441)^{\frac{1}{3}} \approx 3.4297\)

The root is approximately 3.43 to 2 decimal places.