(i) Let \(x = 2 \tan \theta\). Then \(dx = 2 \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(\int_0^2 \frac{8}{(4+x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \frac{8}{(4 + (2 \tan \theta)^2)^2} \cdot 2 \sec^2 \theta \, d\theta.\)

Simplify the expression:

\(4 + (2 \tan \theta)^2 = 4 + 4 \tan^2 \theta = 4 \sec^2 \theta.\)

Thus, the integral becomes:

\(\int_0^{\frac{\pi}{4}} \frac{8 \cdot 2 \sec^2 \theta}{(4 \sec^2 \theta)^2} \, d\theta = \int_0^{\frac{\pi}{4}} \cos^2 \theta \, d\theta.\)

(ii) Use the identity \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\).

Integrate:

\(\int_0^{\frac{\pi}{4}} \cos^2 \theta \, d\theta = \int_0^{\frac{\pi}{4}} \frac{1}{2}(1 + \cos 2\theta) \, d\theta = \frac{1}{2} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_0^{\frac{\pi}{4}}.\)

Evaluate the limits:

\(\frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{2} \cdot 1 - (0 + 0) \right) = \frac{1}{8}(\pi + 2).\)