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

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

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

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

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

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

Let \(u = 1 + 3 \tan x\). Then, \(\frac{du}{dx} = 3 \sec^2 x\), so \(du = 3 \sec^2 x \, dx\).

Rearranging gives \(dx = \frac{du}{3 \sec^2 x}\).

Substitute into the integral:

\(\int \frac{\sqrt{u}}{\cos^2 x} \cdot \frac{du}{3 \sec^2 x} = \int \frac{\sqrt{u}}{3} \, du.\)

This simplifies to:

\(\frac{1}{3} \int u^{\frac{1}{2}} \, du.\)

Integrate:

\(\frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{2}{9} u^{\frac{3}{2}}.\)

Evaluate from \(x = 0\) to \(x = \frac{\pi}{4}\):

When \(x = 0\), \(u = 1 + 3 \tan 0 = 1\).

When \(x = \frac{\pi}{4}\), \(u = 1 + 3 \tan \frac{\pi}{4} = 4\).

Substitute these limits into the integrated function:

\(\frac{2}{9} \left[ 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] = \frac{2}{9} (8 - 1) = \frac{2}{9} \times 7 = \frac{14}{9}.\)

Let \(u = 3x + 1\). Then, \(\frac{du}{dx} = 3\), so \(dx = \frac{1}{3} du\).

Substitute into the integral:

\(\int \frac{3x}{3x+1} \, dx = \int \left( \frac{1}{3} - \frac{1}{3u} \right) du\).

Integrate each term separately:

\(\int \frac{1}{3} \, du = \frac{1}{3} u\).

\(\int -\frac{1}{3u} \, du = -\frac{1}{3} \ln|u|\).

Combine the results:

\(\frac{1}{3} u - \frac{1}{3} \ln|u| + c\).

Substitute back \(u = 3x + 1\):

\(\frac{1}{3}(3x+1) - \frac{1}{3} \ln(3x+1) + c\).

Using the substitution \(x = \sqrt{3} \tan \theta\), we have \(dx = \sqrt{3} \sec^2 \theta \, d\theta\).

The integral becomes:

\(\int \frac{1}{\sqrt{3 + 3 \tan^2 \theta}} \cdot \sqrt{3} \sec^2 \theta \, d\theta = \int \sec \theta \, d\theta.\)

We need to change the limits of integration. When \(x = 1\), \(\sqrt{3} \tan \theta = 1\), so \(\tan \theta = \frac{1}{\sqrt{3}}\), giving \(\theta = \frac{\pi}{6}\).

When \(x = 3\), \(\sqrt{3} \tan \theta = 3\), so \(\tan \theta = \sqrt{3}\), giving \(\theta = \frac{\pi}{3}\).

Thus, the integral becomes:

\(\int_{\pi/6}^{\pi/3} \sec \theta \, d\theta.\)

The integral of \(\sec \theta\) is \(\ln |\sec \theta + \tan \theta|\).

Evaluating from \(\theta = \frac{\pi}{6}\) to \(\theta = \frac{\pi}{3}\):

\(\ln |\sec \frac{\pi}{3} + \tan \frac{\pi}{3}| - \ln |\sec \frac{\pi}{6} + \tan \frac{\pi}{6}|.\)

\(\sec \frac{\pi}{3} = 2\), \(\tan \frac{\pi}{3} = \sqrt{3}\), \(\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}\), \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\).

Thus, the expression becomes:

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

\(= \ln (2 + \sqrt{3}) - \ln \sqrt{3} = \ln \left( \frac{2 + \sqrt{3}}{\sqrt{3}} \right).\)

Let \(u = \sin 4x\), then \(\frac{du}{dx} = 4\cos 4x\) or \(du = 4\cos 4x \, dx\).

Thus, \(dx = \frac{du}{4\cos 4x}\).

Substitute into the integral:

\(\int \cos^3 4x \, dx = \int \cos^3 4x \cdot \frac{du}{4\cos 4x} = \frac{1}{4} \int \cos^2 4x \, du\).

Since \(\cos^2 4x = 1 - \sin^2 4x = 1 - u^2\), the integral becomes:

\(\frac{1}{4} \int (1 - u^2) \, du\).

Integrate to obtain:

\(\frac{1}{4} \left( u - \frac{u^3}{3} \right) + C\).

Evaluate from \(u = \sin 4(0) = 0\) to \(u = \sin 4\left(\frac{1}{24}\pi\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).

\(\frac{1}{4} \left[ \left( \frac{1}{2} - \frac{(\frac{1}{2})^3}{3} \right) - (0 - 0) \right]\).

\(= \frac{1}{4} \left( \frac{1}{2} - \frac{1}{24} \right)\).

\(= \frac{1}{4} \cdot \frac{11}{24}\).

\(= \frac{11}{96}\).