Answer: \(a=\dfrac{9+6\sqrt6}{5}\); \(\dfrac{3+\sqrt3}{4}-\dfrac{\pi}{3}\).

(a) Integrate term by term:

\(\int\left(\frac1x-\frac{1}{2x+3}\right)\,dx =\ln x-\frac12\ln(2x+3).\)

Using the limits \(1\) to \(a\),

\(\left[\ln x-\frac12\ln(2x+3)\right]_1^a=\ln3.\)

So

\(\ln a-\frac12\ln(2a+3)+\frac12\ln5=\ln3.\)

Combining the logarithms gives

\(\ln\left(\frac{a\sqrt5}{\sqrt{2a+3}}\right)=\ln3.\)

Hence

\(\frac{a\sqrt5}{\sqrt{2a+3}}=3.\)

Squaring both sides,

\(5a^2=9(2a+3).\)

Therefore

\(5a^2-18a-27=0.\)

Solving,

\(a=\frac{18\pm\sqrt{18^2+4\cdot5\cdot27}}{10} =\frac{18\pm12\sqrt6}{10} =\frac{9\pm6\sqrt6}{5}.\)

Since \(a\gt 0\),

\(a=\frac{9+6\sqrt6}{5}.\)

(b) An antiderivative is

\(-\frac12\cos\left(2x+\frac{\pi}{3}\right)-x+\frac12\sin2x.\)

Using the limits \(0\) and \(\frac{\pi}{3}\), the value is

\(\left[-\frac12\cos\left(2x+\frac{\pi}{3}\right)-x+\frac12\sin2x\right]_0^{\pi/3}.\)

At \(x=\frac{\pi}{3}\), this is

\(-\frac12\cos\pi-\frac{\pi}{3}+\frac12\sin\frac{2\pi}{3} =\frac12-\frac{\pi}{3}+\frac{\sqrt3}{4}.\)

At \(x=0\), this is

\(-\frac12\cos\frac{\pi}{3}=-\frac14.\)

Therefore the exact value is

\(\frac12-\frac{\pi}{3}+\frac{\sqrt3}{4}+\frac14 =\frac{3+\sqrt3}{4}-\frac{\pi}{3}.\)