(i) Calculate \(f(x) = x^3 - 3x - 7\) at \(x = 2\) and \(x = 3\):

\(f(2) = 2^3 - 3 \times 2 - 7 = 8 - 6 - 7 = -5\)

\(f(3) = 3^3 - 3 \times 3 - 7 = 27 - 9 - 7 = 11\)

Since \(f(2) < 0\) and \(f(3) > 0\), by the Intermediate Value Theorem, \(\alpha\) lies between 2 and 3.

(ii) Using formula \(A\):

\(x_1 = 2.5\)

\(x_2 = (3 \times 2.5 + 7)^{\frac{1}{3}} = 2.4190\)

\(x_3 = (3 \times 2.4190 + 7)^{\frac{1}{3}} = 2.4320\)

\(x_4 = (3 \times 2.4320 + 7)^{\frac{1}{3}} = 2.4268\)

\(x_5 = (3 \times 2.4268 + 7)^{\frac{1}{3}} = 2.4295\)

\(x_6 = (3 \times 2.4295 + 7)^{\frac{1}{3}} = 2.4282\)

\(x_7 = (3 \times 2.4282 + 7)^{\frac{1}{3}} = 2.4288\)

Using formula \(B\):

\(x_1 = 2.5\)

\(x_2 = \frac{2.5^3 - 7}{3} = 2.9583\)

\(x_3 = \frac{2.9583^3 - 7}{3} = 7.3878\)

\(x_4 = \frac{7.3878^3 - 7}{3} = 131.1805\)

Formula \(B\) fails to converge. Formula \(A\) converges to \(\alpha = 2.43\) correct to 2 decimal places.