First, separate the variables:

\(\frac{dy}{y^3} = ke^{-x} dx\)

Integrate both sides:

\(\int y^{-3} \, dy = \int ke^{-x} \, dx\)

This gives:

\(-\frac{1}{2y^2} = -ke^{-x} + c\)

Using the initial condition \(y = 1\) when \(x = 0\):

\(-\frac{1}{2(1)^2} = -k + c\)

\(-\frac{1}{2} = -k + c\)

Using the second condition \(y = \sqrt{e}\) when \(x = 1\):

\(-\frac{1}{2(\sqrt{e})^2} = -ke^{-1} + c\)

\(-\frac{1}{2e} = -\frac{k}{e} + c\)

Solving these equations gives \(k = \frac{1}{2}\) and \(c = 0\).

Substitute back to find \(y\):

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

\(y^2 = e^x\)

\(y = e^{\frac{1}{2}x}\)

(i) To show that \(\frac{d}{d\theta}(\tan^2 \theta) = \frac{2 \tan \theta}{\cos^2 \theta}\), we start by using the chain rule:

\(\frac{d}{d\theta}(\tan^2 \theta) = 2 \tan \theta \cdot \frac{d}{d\theta}(\tan \theta).\)

Since \(\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta = \frac{1}{\cos^2 \theta}\), we have:

\(\frac{d}{d\theta}(\tan^2 \theta) = 2 \tan \theta \cdot \frac{1}{\cos^2 \theta} = \frac{2 \tan \theta}{\cos^2 \theta}.\)

(ii) To solve the differential equation \(x \cos^2 \theta \frac{dx}{d\theta} = 2 \tan \theta + 1\), we separate variables:

\(x \frac{dx}{d\theta} = \frac{2 \tan \theta + 1}{\cos^2 \theta}.\)

Integrating both sides, we have:

\(\int x \, dx = \int \left( \frac{2 \tan \theta}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} \right) d\theta.\)

The left side integrates to \(\frac{1}{2}x^2\).

The right side integrates to \(\tan^2 \theta + \tan \theta\).

Thus, \(\frac{1}{2}x^2 = \tan^2 \theta + \tan \theta + C\).

Using the initial condition \(x = 1\) when \(\theta = \frac{1}{4}\pi\), we find:

\(\frac{1}{2}(1)^2 = \tan^2 \left(\frac{1}{4}\pi\right) + \tan \left(\frac{1}{4}\pi\right) + C.\)

\(\frac{1}{2} = 1 + 1 + C \Rightarrow C = -\frac{3}{2}.\)

Thus, \(\frac{1}{2}x^2 = \tan^2 \theta + \tan \theta - \frac{3}{2}\).

Substituting \(\theta = \frac{1}{3}\pi\), we find:

\(\frac{1}{2}x^2 = \tan^2 \left(\frac{1}{3}\pi\right) + \tan \left(\frac{1}{3}\pi\right) - \frac{3}{2}.\)

\(\frac{1}{2}x^2 = \left(\sqrt{3}\right)^2 + \sqrt{3} - \frac{3}{2} = 3 + \sqrt{3} - \frac{3}{2}.\)

\(\frac{1}{2}x^2 = \frac{3}{2} + \sqrt{3}.\)

\(x^2 = 3 + 2\sqrt{3}.\)

\(x = \sqrt{3 + 2\sqrt{3}} \approx 2.54.\)

Separate the variables: \(\frac{1}{\cos^2 y} dy = 4 \tan x \, dx\).

Integrate both sides: \(\int \sec^2 y \, dy = \int 4 \tan x \, dx\).

The integrals are: \(\tan y = 4 \ln |\sec x| + C\).

Use the initial condition \(x = 0\) and \(y = \frac{1}{4}\pi\):

\(\tan \left( \frac{1}{4}\pi \right) = 4 \ln |\sec 0| + C\).

\(1 = 4 \cdot 0 + C \Rightarrow C = 1\).

The solution is \(\tan y = 4 \ln |\sec x| + 1\).

Substitute \(y = \frac{1}{3}\pi\):

\(\tan \left( \frac{1}{3}\pi \right) = 4 \ln |\sec x| + 1\).

\(\sqrt{3} = 4 \ln |\sec x| + 1\).

\(4 \ln |\sec x| = \sqrt{3} - 1\).

\(\ln |\sec x| = \frac{\sqrt{3} - 1}{4}\).

\(|\sec x| = e^{\frac{\sqrt{3} - 1}{4}}\).

\(\sec x = e^{\frac{\sqrt{3} - 1}{4}}\).

\(x = \arccos \left( \frac{1}{e^{\frac{\sqrt{3} - 1}{4}}} \right)\).

\(x \approx 0.587\).

To solve the differential equation \(\frac{dy}{dx} = e^{-2y} \tan^2 x\), we first separate variables:

\(e^{2y} \, dy = \tan^2 x \, dx\).

Integrate both sides:

\(\int e^{2y} \, dy = \int \tan^2 x \, dx\).

The left side integrates to \(\frac{1}{2} e^{2y}\), and the right side integrates to \(\tan x - x + C\), where \(C\) is the constant of integration.

Thus, \(\frac{1}{2} e^{2y} = \tan x - x + C\).

Using the initial condition \(y = 0\) when \(x = 0\), we find:

\(\frac{1}{2} e^{0} = \tan 0 - 0 + C\).

This gives \(\frac{1}{2} = C\).

Substitute \(C = \frac{1}{2}\) back into the equation:

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

Now, solve for \(y\) when \(x = \frac{1}{4}\pi\):

\(\frac{1}{2} e^{2y} = \tan \frac{1}{4}\pi - \frac{1}{4}\pi + \frac{1}{2}\).

Calculate \(\tan \frac{1}{4}\pi = 1\), so:

\(\frac{1}{2} e^{2y} = 1 - \frac{1}{4}\pi + \frac{1}{2}\).

\(\frac{1}{2} e^{2y} = \frac{3}{2} - \frac{1}{4}\pi\).

\(e^{2y} = 3 - \frac{1}{2}\pi\).

\(2y = \ln(3 - \frac{1}{2}\pi)\).

\(y = \frac{1}{2} \ln(3 - \frac{1}{2}\pi)\).

Calculate \(y \approx 0.179\).

(i) Separate variables: \(\frac{1}{x} dx = \frac{\sin 2\theta}{3 + \cos 2\theta} d\theta\).

Integrate both sides: \(\int \frac{1}{x} dx = \int \frac{\sin 2\theta}{3 + \cos 2\theta} d\theta\).

This gives \(\ln x = -\frac{1}{2} \ln(3 + \cos 2\theta) + C\).

Use initial condition \(x = 3\) when \(\theta = \frac{1}{4}\pi\) to find \(C\):

\(\ln 3 = -\frac{1}{2} \ln(3 + \cos \frac{\pi}{2}) + C\)

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

\(C = \frac{3}{2} \ln 3\)

Thus, \(\ln x = -\frac{1}{2} \ln(3 + \cos 2\theta) + \frac{3}{2} \ln 3\)

Rearrange to find \(x\):

\(x = \sqrt{\frac{27}{3 + \cos 2\theta}}\)

(ii) The least value of \(x\) occurs when \(\cos 2\theta = 1\), giving:

\(x = \sqrt{\frac{27}{3 + 1}} = \frac{3\sqrt{3}}{2}\)