(a) Let \(x = 3 \tan \theta\). Then \(dx = 3 \sec^2 \theta \, d\theta\).

Substitute into the integral:

\(I = \int_0^3 \frac{27}{(9 + x^2)^2} \, dx = \int_0^{\frac{\pi}{4}} \frac{27}{(9 + 9 \tan^2 \theta)^2} \cdot 3 \sec^2 \theta \, d\theta\).

\(= \int_0^{\frac{\pi}{4}} \frac{81 \sec^2 \theta}{81 \sec^4 \theta} \, d\theta\).

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

(b) The integral \(\int \cos^2 \theta \, d\theta\) can be expressed as \(\int \frac{1 + \cos 2\theta}{2} \, d\theta\).

\(= \frac{1}{2} \int (1 + \cos 2\theta) \, d\theta\).

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

\(= \frac{1}{2} \left[ \frac{\pi}{4} + \frac{1}{2} \sin \frac{\pi}{2} - (0 + 0) \right]\).

\(= \frac{1}{2} \left[ \frac{\pi}{4} + \frac{1}{2} \right]\).

\(= \frac{1}{8}(\pi + 2)\).