(a) To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to zero. Using the quotient rule, \(\frac{dy}{dx} = \frac{(9-x)(-\frac{1}{2}x^{-3/2}) - \sqrt{x}}{(9-x)^2 x}\).

Simplifying, \(\frac{dy}{dx} = \frac{-\frac{1}{2}(9-x)x^{-3/2} - x^{-1/2}}{(9-x)^2}\).

Setting \(\frac{dy}{dx} = 0\), we solve for \(x\) to get \(x = 3\).

(b) Using the substitution \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).

The integral becomes \(\int_0^2 \frac{1}{(9-u^2)u} \, 2u \, du = \int_0^2 \frac{2}{9-u^2} \, du\).

This can be integrated using partial fractions or a standard integral formula to obtain \(\frac{1}{3} \ln \left( \frac{3+u}{3-u} \right)\).

Evaluating from \(u = 0\) to \(u = 2\), we get \(\frac{1}{3} \ln 5\).

Let \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u \, du\).

The limits change from \(x = 3\) to \(u = \sqrt{3}\) and \(x = \infty\) to \(u = \infty\).

Substitute into the integral:

\(\int_{\sqrt{3}}^{\infty} \frac{1}{(u^2 + 1)u} \, 2u \, du = \int_{\sqrt{3}}^{\infty} \frac{2}{u^2 + 1} \, du\).

The integral \(\int \frac{2}{u^2 + 1} \, du\) is \(2 \arctan(u)\).

Evaluate from \(\sqrt{3}\) to \(\infty\):

\(2 \left[ \arctan(\infty) - \arctan(\sqrt{3}) \right] = 2 \left[ \frac{\pi}{2} - \frac{\pi}{3} \right]\).

\(= 2 \left[ \frac{\pi}{6} \right] = \frac{1}{3} \pi\).

(i) Let \(x = \cos^2 \theta\). Then \(dx = -2 \cos \theta \sin \theta \, d\theta = -\sin 2\theta \, d\theta\).

Substitute \(x = \cos^2 \theta\) into the integral:

\(\sqrt{\frac{x}{1-x}} = \sqrt{\frac{\cos^2 \theta}{1-\cos^2 \theta}} = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta\).

Thus, the integral becomes:

\(I = \int \cot \theta (-\sin 2\theta \, d\theta) = \int -\cos 2\theta \, d\theta\).

Change the limits: when \(x = \frac{1}{4}\), \(\cos^2 \theta = \frac{1}{4}\), so \(\theta = \frac{1}{3}\pi\). When \(x = \frac{3}{4}\), \(\cos^2 \theta = \frac{3}{4}\), so \(\theta = \frac{1}{6}\pi\).

Therefore, \(I = \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} 2 \cos^2 \theta \, d\theta\).

(ii) The integral \(\int 2 \cos^2 \theta \, d\theta = \int (1 + \cos 2\theta) \, d\theta = \theta + \frac{1}{2} \sin 2\theta\).

Evaluate from \(\frac{1}{6}\pi\) to \(\frac{1}{3}\pi\):

\(\left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} = \left( \frac{1}{3}\pi + \frac{1}{2} \sin \frac{2}{3}\pi \right) - \left( \frac{1}{6}\pi + \frac{1}{2} \sin \frac{1}{3}\pi \right)\).

\(= \frac{1}{3}\pi - \frac{1}{6}\pi = \frac{1}{6}\pi\).

Let \(u = \sqrt{x}\), then \(x = u^2\).

Differentiate to find \(dx\):

\(\frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2u}\)

Thus, \(du = \frac{1}{2u} \, dx\) or \(dx = 2u \, du\).

Substitute \(x = u^2\) and \(dx = 2u \, du\) into the integral:

\(I = \int_{1}^{4} \frac{(u) - 1}{2(u^2 + u)} \, 2u \, du\)

\(I = \int_{1}^{4} \frac{u(u - 1)}{2(u^2 + u)} \, 2u \, du\)

\(I = \int_{1}^{4} \frac{u(u - 1)}{u^2 + u} \, du\)

Change the limits of integration:

When \(x = 1\), \(u = \sqrt{1} = 1\).

When \(x = 4\), \(u = \sqrt{4} = 2\).

Thus, \(I = \int_{1}^{2} \frac{u - 1}{u + 1} \, du\).

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

Substitute \(x^2 = u - 1\) into the integral:

\(I = \int_0^1 \frac{x^5}{(1+x^2)^3} \, dx = \int_1^2 \frac{(u-1)^{5/2}}{u^3} \cdot \frac{du}{2\sqrt{u-1}}\).

Simplifying, we get:

\(I = \int_1^2 \frac{(u-1)^2}{2u^3} \, du\).

(ii) To find the exact value of \(I\), integrate:

\(\int \frac{(u-1)^2}{2u^3} \, du = \int \left( \frac{1}{2u} - \frac{1}{u^2} + \frac{1}{4u^2} \right) \, du\).

This gives:

\(\frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2}\).

Evaluate from 1 to 2:

\(\left[ \frac{1}{2} \ln u + \frac{1}{u} - \frac{1}{4u^2} \right]_1^2 = \left( \frac{1}{2} \ln 2 + \frac{1}{2} - \frac{1}{16} \right) - \left( 0 + 1 - \frac{1}{4} \right)\).

Simplifying, we find:

\(\frac{1}{2} \ln 2 - \frac{5}{16}\).