Answer: (a) \(\frac43\). (b) \(\frac{3\sqrt3-5}{2}\).

(a)

\(\frac{\sqrt{128}}{\sqrt{72}}=\sqrt{\frac{128}{72}}=\sqrt{\frac{16}{9}}=\frac43.\)

(b) First rationalise each term:

\(\frac{1}{1+\sqrt3}=\frac{1-\sqrt3}{1-3}=\frac{\sqrt3-1}{2}.\)

Also

\(\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3(3-2\sqrt3)}{(3+2\sqrt3)(3-2\sqrt3)}.\)

The denominator is \(9-12=-3\), so

\(\frac{\sqrt3}{3+2\sqrt3}=\frac{3\sqrt3-6}{-3}=2-\sqrt3.\)

Therefore

\(\frac{1}{1+\sqrt3}-\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3-1}{2}-(2-\sqrt3).\)

Putting this over a common denominator gives

\(\frac{\sqrt3-1-4+2\sqrt3}{2}=\frac{3\sqrt3-5}{2}.\)

Hence

\(\boxed{\frac{3\sqrt3-5}{2}}.\)