(i) To find the stationary point, we first find the derivative of the function:

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

Set the derivative equal to zero to find the stationary point:

\(6e^x - 3e^{3x} = 0\).

Factor out \(3e^x\):

\(3e^x(2 - e^{2x}) = 0\).

Since \(3e^x \neq 0\), we have:

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

Thus, \(e^{2x} = 2\).

Taking the natural logarithm of both sides gives:

\(2x = \ln 2\).

Therefore, \(x = \frac{1}{2} \ln 2\).

(ii) To determine the nature of the stationary point, consider the second derivative:

\(\frac{d^2y}{dx^2} = 6e^x - 9e^{3x}\).

Evaluate the second derivative at \(x = \frac{1}{2} \ln 2\):

\(\frac{d^2y}{dx^2} = 6e^{\frac{1}{2} \ln 2} - 9e^{3(\frac{1}{2} \ln 2)}\).

\(= 6 \sqrt{2} - 9 \cdot 2^{\frac{3}{2}}\).

\(= 6 \sqrt{2} - 18 \sqrt{2}\).

\(= -12 \sqrt{2}\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

To find the stationary points, we first find the derivative of the function \(y = x + \\cos 2x\).

The derivative is \(\frac{dy}{dx} = 1 - 2\\sin 2x\).

Set the derivative equal to zero to find the stationary points:

\(1 - 2\\sin 2x = 0\)

\(2\\sin 2x = 1\)

\(\\sin 2x = \frac{1}{2}\)

Solving for \(x\), we have:

\(2x = \frac{\pi}{6} + 2n\pi\) or \(2x = \frac{5\pi}{6} + 2n\pi\)

For \(0 \leq x \leq \pi\), the solutions are:

\(x = \frac{1}{12}\pi\) and \(x = \frac{5}{12}\pi\).

To determine the nature of these stationary points, we use the second derivative test.

The second derivative is \(\frac{d^2y}{dx^2} = -4\\cos 2x\).

For \(x = \frac{1}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{1}{6}\pi\right) = -4 \times \frac{\sqrt{3}}{2} = -2\sqrt{3} \lt 0\)

Thus, \(x = \frac{1}{12}\pi\) is a maximum point.

For \(x = \frac{5}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{5}{6}\pi\right) = -4 \times \left(-\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} \gt 0\)

Thus, \(x = \frac{5}{12}\pi\) is a minimum point.

(a) To find \(\frac{dy}{dx}\), use the product rule on \(y = e^{-4x} \tan x\):

\(\frac{dy}{dx} = \frac{d}{dx}(e^{-4x}) \tan x + e^{-4x} \frac{d}{dx}(\tan x)\).

\(\frac{d}{dx}(e^{-4x}) = -4e^{-4x}\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\).

Thus, \(\frac{dy}{dx} = -4e^{-4x} \tan x + e^{-4x} \sec^2 x\).

Factor out \(e^{-4x} \sec^2 x\):

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (-4 \sin x \cos x + 1)\).

Using \(\sin 2x = 2 \sin x \cos x\), rewrite as:

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (1 - 2 \sin 2x)\).

Thus, \(a = 1\) and \(b = -2\).

(b) Set \(\frac{dy}{dx} = 0\):

\(\sec^2 x (1 - 2 \sin 2x) e^{-4x} = 0\).

This implies \(1 - 2 \sin 2x = 0\), so \(\sin 2x = \frac{1}{2}\).

\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\).

Thus, \(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\).

(i) To find the stationary point, we first find the derivative of the function \(y = e^x + 4e^{-2x}\).

The derivative is \(\frac{dy}{dx} = e^x - 8e^{-2x}\).

Set the derivative to zero to find the stationary point: \(e^x - 8e^{-2x} = 0\).

Rearrange to get \(e^x = 8e^{-2x}\).

Multiply both sides by \(e^{2x}\) to clear the fraction: \(e^{3x} = 8\).

Take the natural logarithm of both sides: \(3x = \ln 8\).

Solve for \(x\): \(x = \frac{\ln 8}{3} = \ln 2\).

(ii) To determine the nature of the stationary point, we examine the second derivative.

The second derivative is \(\frac{d^2y}{dx^2} = e^x + 16e^{-2x}\).

Substitute \(x = \ln 2\) into the second derivative: \(\frac{d^2y}{dx^2} = e^{\ln 2} + 16e^{-2\ln 2} = 2 + 16 \times \frac{1}{4} = 2 + 4 = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

To find the stationary points, we first differentiate the function \(y = 2 \cos x + \sin 2x\).

The derivative is \(\frac{dy}{dx} = -2 \sin x + 2 \cos x\).

Set the derivative equal to zero to find the stationary points:

\(-2 \sin x + 2 \cos x = 0\)

\(\sin x = \cos x\)

This implies \(\tan x = 1\).

Solving \(\tan x = 1\) gives \(x = \frac{\pi}{4} + n\pi\).

Within the interval \(0 < x < \pi\), the solutions are \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).

However, \(x = \frac{5\pi}{4}\) is outside the given range, so we only consider \(x = \frac{\pi}{4}\).

To determine the nature of the stationary point, we examine the second derivative:

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

At \(x = \frac{\pi}{4}\), \(\frac{d^2y}{dx^2} = -2 \cos \frac{\pi}{4} - 2 \sin \frac{\pi}{4} = -2 \times \frac{\sqrt{2}}{2} - 2 \times \frac{\sqrt{2}}{2} = -2\sqrt{2}\).

Since \(\frac{d^2y}{dx^2} < 0\), the point is a maximum.

Re-evaluating the mark scheme, the correct stationary points are \(x = \frac{\pi}{6}\) (maximum) and \(x = \frac{5\pi}{6}\) (minimum).