(i) The derivatives of the curves are \(2e^{2x-3}\) and \(\frac{2}{x}\). Setting these equal for parallel tangents gives:

\(2e^{2a-3} = \frac{2}{a}\)

\(e^{2a-3} = \frac{1}{a}\)

Taking natural logarithms on both sides:

\(2a - 3 = -\ln a\)

\(2a = 3 - \ln a\)

\(a = \frac{1}{2}(3 - \ln a)\)

(ii) Consider \(f(a) = a - \frac{1}{2}(3 - \ln a)\). Calculate \(f(1)\) and \(f(2)\):

\(f(1) = 1 - \frac{1}{2}(3 - \ln 1) = 1 - \frac{3}{2} = -0.5\)

\(f(2) = 2 - \frac{1}{2}(3 - \ln 2) = 2 - \frac{1}{2}(3 - 0.693) = 0.1535\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(iii) Using the iterative formula \(a_{n+1} = \frac{1}{2}(3 - \ln a_n)\):

Start with \(a_1 = 1.5\).

\(a_2 = \frac{1}{2}(3 - \ln 1.5) = 1.3466\)

\(a_3 = \frac{1}{2}(3 - \ln 1.3466) = 1.3539\)

\(a_4 = \frac{1}{2}(3 - \ln 1.3539) = 1.3519\)

\(a_5 = \frac{1}{2}(3 - \ln 1.3519) = 1.3525\)

\(a_6 = \frac{1}{2}(3 - \ln 1.3525) = 1.3522\)

\(a_7 = \frac{1}{2}(3 - \ln 1.3522) = 1.3523\)

The value of \(a\) correct to 2 decimal places is 1.35.

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = e^{-\frac{1}{2}x^2} \sqrt{(1 + 2x^2)}\). Differentiate \(y\) with respect to \(x\) using the product and chain rules. Set the derivative to zero and solve for \(x\). The solution is \(x = \frac{1}{\sqrt{2}}\).

(ii) The iterative formula \(x_{n+1} = \sqrt{(\ln(4 + 8x_n^2))}\) converges to \(\alpha\). Therefore, \(\alpha = \sqrt{(\ln(4 + 8\alpha^2))}\). Rearrange to find \(e^{\alpha^2} = 4 + 8\alpha^2\). Substitute \(y = 0.5\) into the original curve equation and show that \(\alpha\) satisfies this equation.

(iii) Use the iterative formula starting with \(x_1 = 2\) and calculate successive values until the result stabilizes to 2 decimal places. The iterations yield \(\alpha = 1.86\).

(i) To find the stationary point, we first find the derivative of \(y = \frac{\ln x}{x + 1}\) using the quotient rule:

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

Setting the derivative to zero gives:

\(1 - \ln x = 0\) or \(\ln x = 1\).

Solving \(\ln x = 1\) gives \(x = e\), but we need to verify the given equation:

Rearrange \(x = \frac{x + 1}{\ln x}\) to \(x \ln x = x + 1\), which simplifies to \(\ln x = 1 + \frac{1}{x}\).

Check the interval \(x = 3\) and \(x = 4\):

For \(x = 3\), \(\ln 3 \approx 1.0986\) and \(\frac{3 + 1}{3} \approx 1.3333\).

For \(x = 4\), \(\ln 4 \approx 1.3863\) and \(\frac{4 + 1}{4} = 1.25\).

Since \(\ln x\) changes from greater than \(\frac{x + 1}{x}\) to less than \(\frac{x + 1}{x}\) between 3 and 4, there is a root in this interval.

(ii) Using the iterative formula \(x_{n+1} = \frac{x_n + 1}{\ln x_n}\):

Start with \(x_1 = 3.5\):

\(x_2 = \frac{3.5 + 1}{\ln 3.5} \approx 3.5718\)

\(x_3 = \frac{3.5718 + 1}{\ln 3.5718} \approx 3.5911\)

\(x_4 = \frac{3.5911 + 1}{\ln 3.5911} \approx 3.5900\)

\(x_5 = \frac{3.5900 + 1}{\ln 3.5900} \approx 3.5900\)

The \(x\)-coordinate correct to 2 decimal places is 3.59.

(i) To find the \(x\)-coordinate of the minimum point \(M\), we need to find the derivative of \(y = \frac{\sin x}{x}\) and set it to zero.

Using the quotient rule, the derivative is:

\(y' = \frac{x \cos x - \sin x}{x^2}\).

Setting \(y' = 0\), we have:

\(x \cos x - \sin x = 0\).

Rearranging gives:

\(x \cos x = \sin x\).

Dividing both sides by \(\cos x\), we get:

\(x = \tan x\).

(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n) + \pi\), we start with an initial guess and iterate:

1. \(x_1 = \arctan(x_0) + \pi\)

2. \(x_2 = \arctan(x_1) + \pi\)

Continue iterating until the value stabilizes to 4 decimal places.

The final \(x\)-coordinate of \(M\) is 4.49, correct to 2 decimal places.

(i) To find the stationary point, we first find the derivative of \(y = \ln x + \frac{2}{x}\). The derivative is \(\frac{dy}{dx} = \frac{1}{x} - \frac{2}{x^2}\).

Setting \(\frac{dy}{dx} = 0\), we have \(\frac{1}{x} - \frac{2}{x^2} = 0\).

Multiplying through by \(x^2\), we get \(x - 2 = 0\), so \(x = 2\).

Substituting \(x = 2\) back into the original equation, \(y = \ln 2 + \frac{2}{2} = \ln 2 + 1\).

To determine if it is a minimum or maximum, consider the second derivative: \(\frac{d^2y}{dx^2} = -\frac{1}{x^2} + \frac{4}{x^3}\).

At \(x = 2\), \(\frac{d^2y}{dx^2} = -\frac{1}{4} + \frac{4}{8} = \frac{1}{4} > 0\), indicating a minimum point.

(ii) The iterative formula is \(x_{n+1} = \frac{2}{3 - \ln x_n}\). As \(n \to \infty\), \(x_n \to \alpha\), so \(\alpha = \frac{2}{3 - \ln \alpha}\).

Rearranging gives \(3 = \ln \alpha + \frac{2}{\alpha}\), which matches the equation \(y = 3\).

(iii) Using the iterative formula:

\(x_1 = 1\)

\(x_2 = \frac{2}{3 - \ln 1} = \frac{2}{3} \approx 0.67\)

\(x_3 = \frac{2}{3 - \ln 0.67} \approx 0.58\)

\(x_4 = \frac{2}{3 - \ln 0.58} \approx 0.56\)

\(x_5 = \frac{2}{3 - \ln 0.56} \approx 0.56\)

Thus, \(\alpha \approx 0.56\) to 2 decimal places.