Answer: (a) \(4\sin\dfrac{\pi}{8}-4\sin\dfrac{\pi}{12}=0.495\); (b) \(\dfrac14\ln(4x-3)-\dfrac{1}{2x^2}+C\).

We start with the main method. Integrate term by term, remembering that an integral of the form \(1/(ax+b)\) gives a logarithm with factor \(1/a\).

For part (a),

\(\int \cos\frac{x}{4}\,dx=4\sin\frac{x}{4}.\)

Therefore

\(\int_{\pi/3}^{\pi/2}\cos\frac{x}{4}\,dx =\left[4\sin\frac{x}{4}\right]_{\pi/3}^{\pi/2}.\)

This gives

\(4\sin\frac{\pi}{8}-4\sin\frac{\pi}{12}=0.495457\ldots\)

So the value is \(0.495\) to 3 significant figures.

For part (b), integrate each term separately:

\(\int\frac{1}{4x-3}\,dx=\frac14\ln(4x-3)\)

and

\(\int x^{-3}\,dx=\frac{x^{-2}}{-2}=-\frac{1}{2x^2}.\)

Hence

\(\int\left(\frac{1}{4x-3}+\frac{1}{x^3}\right)\,dx =\frac14\ln(4x-3)-\frac{1}{2x^2}+C.\)

This completes the solution and gives the required result.