$2+\sqrt 3=(\frac{1+\sqrt 3)^2}{2}$, therefore $\sqrt{2+\sqrt 3}=\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}.$

Squaring we have $a+b+2\sqrt{ab}=2+\sqrt{3}$. So if we let $x\in\mathbb{Q}$ be equal to $2-a-b$ then $2\sqrt{ab}=x+\sqrt{3}$. Again if we square we get $4ab=x^2+3+2x\sqrt{3}$, but since $a,b$ are rational, so is the $RHS$ which means $x\sqrt{3}$ is rational. So $x=0\implies a+b=2$. So $2+2\sqrt{ab}=a+b+\sqrt{3}=2+\sqrt{3}$. This means $2\sqrt{ab}=\sqrt{3}$ i.e. $ab=\frac{3}{4}$. Solving the equations $a+b=2$ and $ab=\frac{3}{4}$ yields the solutions $\left(\frac{1}{2},\frac{3}{2}\right)$ and $\left(\frac{3}{2},\frac{1}{2}\right)$.

This question can be solved in following way which is known for. More that 150 years can be found on gs carr pure and applied mathematics so the trick is \[ sqrt{a+\sqrtb] \= √x+√y for some rational numbers x, y now assume from problem that a+b=2 and a-b=√(2^2-3) now on solving we get a=3/2 and b=1/2 so they are required answers