(i) Differentiate \(y = \frac{1 + e^{-x}}{1 - e^{-x}}\) using the quotient rule:

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

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

Since \(e^{-x} > 0\) for \(x > 0\), \(-2e^{-x} < 0\). Therefore, \(\frac{dy}{dx}\) is always negative.

(ii) Set \(\frac{dy}{dx} = -1\):

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

Simplify to \(2e^{-a} = (1 - e^{-a})^2\)

Let \(u = e^{-a}\), then \(2u = (1 - u)^2\)

Expand and simplify: \(2u = 1 - 2u + u^2\)

Rearrange to \(u^2 - 4u + 1 = 0\)

Solving this quadratic equation gives \(u = 2 \pm \sqrt{3}\)

Since \(u = e^{-a} > 0\), choose \(u = 2 - \sqrt{3}\)

Thus, \(e^{-a} = 2 - \sqrt{3}\)

\(a = -\ln(2 - \sqrt{3}) = \ln(2 + \sqrt{3})\)

To find the coordinates where the gradient of the tangent is \(\frac{1}{4}\), we first need to find the derivative of \(y = \frac{x}{1 + \ln x}\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x\) and \(v = 1 + \ln x\).

Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{x}\).

So, \(\frac{dy}{dx} = \frac{(1 + \ln x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{1 + \ln x - 1}{(1 + \ln x)^2} = \frac{\ln x}{(1 + \ln x)^2}\).

Set \(\frac{dy}{dx} = \frac{1}{4}\):

\(\frac{\ln x}{(1 + \ln x)^2} = \frac{1}{4}\).

Cross-multiply to get:

\(4 \ln x = (1 + \ln x)^2\).

Expand and rearrange to form a quadratic equation:

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

Solving this quadratic equation, we find \(\ln x = 1\), so \(x = e\).

Substitute \(x = e\) back into the original equation to find \(y\):

\(y = \frac{e}{1 + \ln e} = \frac{e}{1 + 1} = \frac{e}{2}\).

Thus, the exact coordinates are \((e, \frac{1}{2}e)\).

(i) Let \(u = 1 + 3 \cos^2 x\). Then \(y = \frac{2}{3} \ln(u)\).

Using the chain rule, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{1}{u} \cdot \frac{du}{dx}\).

We have \(\frac{du}{dx} = -6 \cos x \sin x = -3 \sin(2x)\).

Thus, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{-3 \sin(2x)}{1 + 3 \cos^2 x}\).

Using \(\sin(2x) = \frac{2 \tan x}{1 + \tan^2 x}\) and \(\cos^2 x = \frac{1}{1 + \tan^2 x}\), we get:

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

(ii) Set \(\frac{dy}{dx} = -1\):

\(\frac{-4 \tan x}{4 + \tan^2 x} = -1\).

Solving, \(4 \tan x = 4 + \tan^2 x\).

\(\tan^2 x - 4 \tan x + 4 = 0\).

Solving the quadratic equation, \(\tan x = 2 \pm \sqrt{0} = 2\).

Thus, \(x = \arctan(2) \approx 1.11\).

To find the point where the tangent is parallel to the \(x\)-axis, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-ax} \tan x\) is:

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

\(\frac{dy}{dx} = e^{-ax} \sec^2 x - ae^{-ax} \tan x\).

Setting \(\frac{dy}{dx} = 0\):

\(e^{-ax} \sec^2 x - ae^{-ax} \tan x = 0\).

\(\sec^2 x = a \tan x\).

Using \(\sec^2 x = 1 + \tan^2 x\), we have:

\(1 + \tan^2 x = a \tan x\).

Rearranging gives the quadratic equation:

\(\tan^2 x - a \tan x + 1 = 0\).

For the quadratic to have only one root, the discriminant must be zero:

\(b^2 - 4ac = 0\).

\((-a)^2 - 4 \cdot 1 \cdot 1 = 0\).

\(a^2 - 4 = 0\).

\(a^2 = 4\).

\(a = 2\) (since \(a\) is positive).

Substitute \(a = 2\) back into the quadratic equation:

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

\((\tan x - 1)^2 = 0\).

\(\tan x = 1\).

\(x = \frac{\pi}{4}\) (within the interval \(0 < x < \frac{1}{2}\pi\)).

Thus, \(a = 2\) and \(x = \frac{1}{4}\pi\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = e^{-2x}\) and \(v = \tan x\).

Then \(u' = -2e^{-2x}\) and \(v' = \sec^2 x\).

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

Factor out \(e^{-2x}\): \(\frac{dy}{dx} = e^{-2x}(-2 \tan x + \sec^2 x)\).

Rewrite as \(e^{-2x}(-2 + \sec^2 x)\) using \(\sec^2 x = 1 + \tan^2 x\).

Thus, \(\frac{dy}{dx} = e^{-2x}(-2 + 1 + \tan^2 x) = e^{-2x}(a + b \tan x)^2\) where \(a = 1\) and \(b = 1\).

(ii) The gradient \(\frac{dy}{dx} = e^{-2x}(-2 + \sec^2 x)\) is never negative because \(\sec^2 x \geq 1\), making \(-2 + \sec^2 x \geq -1\).

(iii) The gradient is least when \(\sec^2 x\) is minimized, which occurs when \(x = \frac{1}{4} \pi\).