To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the quotient rule, the derivative is:

\(\frac{dy}{dx} = \frac{6x(1-x^2)e^{3x^2-1} + 2xe^{3x^2-1}}{(1-x^2)^2}\).

Simplifying the numerator, we have:

\(6x(1-x^2) + 2x = 6x - 6x^3 + 2x = 8x - 6x^3\).

Setting the numerator equal to zero gives:

\(8x - 6x^3 = 0\).

Factoring out \(2x\), we get:

\(2x(4 - 3x^2) = 0\).

This gives solutions \(x = 0\) or \(4 - 3x^2 = 0\).

Solving \(4 - 3x^2 = 0\) gives \(x^2 = \frac{4}{3}\), so \(x = \pm \frac{2\sqrt{3}}{3}\).

Substituting these \(x\)-values back into the original equation to find \(y\):

For \(x = 0\), \(y = e^{-1}\).

For \(x = \pm \frac{2\sqrt{3}}{3}\), \(y = -3e^3\).

Thus, the stationary points are \((0, e^{-1})\), \(\left( \frac{2\sqrt{3}}{3}, -3e^3 \right)\), and \(\left( -\frac{2\sqrt{3}}{3}, -3e^3 \right)\).

(a) To find the stationary point, we need to differentiate \(y = e^{2x}(\sin x + 3 \cos x)\) and set the derivative to zero.

Using the product rule, the derivative is:

\(y' = e^{2x}(2(\sin x + 3 \cos x)) + e^{2x}(\cos x - 3 \sin x)\)

Simplifying, we get:

\(y' = e^{2x}(2 \sin x + 6 \cos x + \cos x - 3 \sin x)\)

\(y' = e^{2x}(-\sin x + 7 \cos x)\)

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

\(-\sin x + 7 \cos x = 0\)

\(\tan x = 7\)

Solving for \(x\) in the interval \(0 \leq x \leq \pi\), we find:

\(x = 1.43\) (to 2 decimal places)

(b) To determine the nature of the stationary point, we check the sign of \(y'\) around \(x = 1.43\).

For \(x = 1.42\), \(y' = 0.06e^{2.84} > 0\)

For \(x = 1.44\), \(y' = -0.07e^{2.88} < 0\)

Since \(y'\) changes from positive to negative, the stationary point at \(x = 1.43\) is a maximum point.

To find the derivative \(\frac{dy}{dx}\) of \(y = \frac{e^{-2x}}{1-x^2}\), we use the quotient rule:

\(\frac{dy}{dx} = \frac{(1-x^2)(-2e^{-2x}) - e^{-2x}(-2x)}{(1-x^2)^2}\)

Simplifying, we get:

\(\frac{dy}{dx} = \frac{-2e^{-2x}(1-x^2) + 2xe^{-2x}}{(1-x^2)^2}\)

To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(-2e^{-2x}(1-x^2) + 2xe^{-2x} = 0\)

Factor out \(e^{-2x}\):

\(e^{-2x}(-2(1-x^2) + 2x) = 0\)

Since \(e^{-2x} \neq 0\), solve:

\(-2 + 2x^2 + 2x = 0\)

Rearrange to form a quadratic equation:

\(2x^2 + 2x - 2 = 0\)

Divide by 2:

\(x^2 + x - 1 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 1, c = -1\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\)

\(x = \frac{-1 \pm \sqrt{5}}{2}\)

Calculate the positive root within the interval \(-1 < x < 1\):

\(x = \frac{-1 + \sqrt{5}}{2} \approx 0.618\)

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the quotient rule for differentiation, \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{3x}\) and \(v = \tan \frac{1}{2}x\).

First, find \(\frac{du}{dx} = 3e^{3x}\) and \(\frac{dv}{dx} = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Substitute into the quotient rule:

\(\frac{dy}{dx} = \frac{\tan \frac{1}{2}x \cdot 3e^{3x} - e^{3x} \cdot \frac{1}{2} \sec^2 \frac{1}{2}x}{\tan^2 \frac{1}{2}x}\).

Simplify and set \(\frac{dy}{dx} = 0\):

\(3\tan \frac{1}{2}x = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Rearrange to form a quadratic in \(\tan \frac{1}{2}x\):

\(6\tan^2 \frac{1}{2}x = \sec^2 \frac{1}{2}x\).

Using \(\sec^2 \theta = 1 + \tan^2 \theta\), substitute:

\(6\tan^2 \frac{1}{2}x = 1 + \tan^2 \frac{1}{2}x\).

Solve for \(\tan \frac{1}{2}x\):

\(5\tan^2 \frac{1}{2}x = 1\).

\(\tan \frac{1}{2}x = \pm \frac{1}{\sqrt{5}}\).

Find \(x\) in the interval \(0 < x < \pi\):

\(x = 2 \arctan \left( \frac{1}{\sqrt{5}} \right) \approx 0.340\).

\(x = 2 \arctan \left( -\frac{1}{\sqrt{5}} \right) + \pi \approx 2.802\).

Thus, the \(x\)-coordinates of the stationary points are \(0.340\) and \(2.802\).

(i) To find the stationary point, we need to differentiate \(y = \frac{2 - \\sin x}{\\cos x}\).

Rewrite the function as \(y = 2\\sec x - \\tan x\).

The derivative is \(y' = 2\\sec x \\tan x - \\sec^2 x\).

Set the derivative to zero: \(2\\sec x \\tan x - \\sec^2 x = 0\).

Factor to get \(\\sec x (2\\tan x - \\sec x) = 0\).

Since \(\\sec x \neq 0\), solve \(2\\tan x = \\sec x\).

Using \(\\tan x = \\frac{\\sin x}{\\cos x}\) and \(\\sec x = \\frac{1}{\\cos x}\), we have \(2\\frac{\\sin x}{\\cos x} = \\frac{1}{\\cos x}\).

Simplify to \(2\\sin x = 1\), giving \(\\sin x = \\frac{1}{2}\).

In the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), \(x = \frac{1}{6}\pi\).

Substitute back to find \(y\): \(y = \frac{2 - \\sin(\frac{1}{6}\pi)}{\\cos(\frac{1}{6}\pi)} = \sqrt{3}\).

Thus, the coordinates are \(\left( \frac{1}{6}\pi, \sqrt{3} \right)\).

(ii) To determine the nature of the stationary point, consider the second derivative or test values around \(x = \frac{1}{6}\pi\).

Since the mark scheme indicates it is a minimum, we conclude it is a minimum point.