(i) Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{2}{3} \left( 1 + \frac{1}{1^2} \right) = \frac{2}{3} \times 2 = \frac{4}{3} \approx 1.3333\).

Calculate \(x_3 = \frac{2}{3} \left( 1.3333 + \frac{1}{(1.3333)^2} \right) \approx 1.2639\).

Calculate \(x_4 = \frac{2}{3} \left( 1.2639 + \frac{1}{(1.2639)^2} \right) \approx 1.2600\).

Calculate \(x_5 = \frac{2}{3} \left( 1.2600 + \frac{1}{(1.2600)^2} \right) \approx 1.2599\).

Since the values are converging, \(\alpha \approx 1.26\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\).

Solving \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\) gives \(\alpha^3 = 2\).

Thus, \(\alpha = \sqrt[3]{2}\) or \(2^{1/3}\).

(a) To show that \(p\) satisfies \(\tan p = \frac{1}{1+p}\), start by equating the functions at \(x = p\):

\(\cos p = \frac{k}{1+p}\).

Differentiate both functions:

\(\frac{d}{dx}(\cos x) = -\sin x\)

\(\frac{d}{dx}\left(\frac{k}{1+x}\right) = -\frac{k}{(1+x)^2}\).

Equate the derivatives at \(x = p\):

\(-\sin p = -\frac{k}{(1+p)^2}\).

Substitute \(k = (1+p)\cos p\) into the derivative equation:

\(-\sin p = -\frac{(1+p)\cos p}{(1+p)^2}\).

Simplify to obtain:

\(\tan p = \frac{1}{1+p}\).

(b) Use the iterative formula:

\(p_{n+1} = \arctan\left(\frac{1}{1+p_n}\right)\).

Start with an initial guess, say \(p_0 = 0.5\), and iterate:

\(p_1 = \arctan\left(\frac{1}{1+0.5}\right) = 0.54042\)

\(p_2 = \arctan\left(\frac{1}{1+0.54042}\right) = 0.56609\)

\(p_3 = \arctan\left(\frac{1}{1+0.56609}\right) = 0.56777\)

\(p_4 = \arctan\left(\frac{1}{1+0.56777}\right) = 0.56799\)

\(p_5 = \arctan\left(\frac{1}{1+0.56799}\right) = 0.56800\)

The value of \(p\) correct to 3 decimal places is \(0.568\).

(c) Substitute \(p = 0.568\) back into \(\cos p = \frac{k}{1+p}\):

\(\cos 0.568 = \frac{k}{1+0.568}\).

Calculate \(\cos 0.568 \approx 0.842\).

\(0.842 = \frac{k}{1.568}\).

\(k = 0.842 \times 1.568 \approx 1.32\).

(i) To find the value of \(b\), we need to find the integer root of the polynomial \(x^4 - 2x^3 - 7x - 6 = 0\). By inspection or using the Rational Root Theorem, we find that \(b = 3\) is a root.

(ii) To show that \(a\) satisfies the equation \(a = -\frac{1}{3}(2 + a^2 + a^3)\), perform polynomial division of \(x^4 - 2x^3 - 7x - 6\) by \(x - 3\). The quotient is \(x^3 + x^2 + 3x + 2\). Setting this equal to zero gives \(x^3 + x^2 + 3x + 2 = 0\). Rearranging gives \(a = -\frac{1}{3}(2 + a^2 + a^3)\).

(iii) Using the iterative formula \(a_{n+1} = -\frac{1}{3}(2 + a_n^2 + a_n^3)\), start with an initial guess, say \(a_0 = -1\). Calculate successive iterations:

\(a_1 = -\frac{1}{3}(2 + (-1)^2 + (-1)^3) = -0.66667\)

\(a_2 = -\frac{1}{3}(2 + (-0.66667)^2 + (-0.66667)^3) = -0.71605\)

\(a_3 = -\frac{1}{3}(2 + (-0.71605)^2 + (-0.71605)^3) = -0.71484\)

\(a_4 = -\frac{1}{3}(2 + (-0.71484)^2 + (-0.71484)^3) = -0.71506\)

\(a_5 = -\frac{1}{3}(2 + (-0.71506)^2 + (-0.71506)^3) = -0.71498\)

Continue until the value stabilizes to \(a \approx -0.715\) to 3 decimal places.

(i) Start with \(x_1 = 2\).

Calculate \(x_2 = \frac{2(2)^6 + 12(2)}{3(2)^5 + 8} = \frac{128 + 24}{96 + 8} = \frac{152}{104} = 1.461538\).

Calculate \(x_3 = \frac{2(1.461538)^6 + 12(1.461538)}{3(1.461538)^5 + 8} \approx 1.336405\).

Calculate \(x_4 = \frac{2(1.336405)^6 + 12(1.336405)}{3(1.336405)^5 + 8} \approx 1.320756\).

Calculate \(x_5 = \frac{2(1.320756)^6 + 12(1.320756)}{3(1.320756)^5 + 8} \approx 1.319545\).

Calculate \(x_6 = \frac{2(1.319545)^6 + 12(1.319545)}{3(1.319545)^5 + 8} \approx 1.319505\).

Calculate \(x_7 = \frac{2(1.319505)^6 + 12(1.319505)}{3(1.319505)^5 + 8} \approx 1.319505\).

Thus, \(\alpha \approx 1.3195\) to 4 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\).

Solving \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\) gives \(\alpha^5 = 4\).

Thus, \(\alpha = \sqrt[5]{4}\).

(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.