(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:

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

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

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

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

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