jmc

First, we simplify $\sqrt{21+12\sqrt{3}}.$ Let $$\sqrt{21+12\sqrt{3}}=x+y.$$Squaring both sides, we get $$21+12\sqrt{3}=x^2+2xy+y^2.$$To make the right-hand side look like the left-hand side, we set $x^2+y^2=21$ and $2xy=12\sqrt{3},$ so $xy=6\sqrt{3}.$ Then $x^2{y}^2=108,$ so by Vieta's formulas, $x^2$ and $y^2$ are the roots of the quadratic $$t^2-21t+108=0.$$This factors as $(t-9)(t-12)=0,$ whose solutions are 9 and 12. Therefore, $$\sqrt{21+12\sqrt{3}}=\sqrt{9}+\sqrt{12}=3+2\sqrt{3}.$$Now we must simplify $$\sqrt{1+3+2\sqrt{3}}=\sqrt{4+2\sqrt{3}}.$$Performing the same technique gives us $$\sqrt{4+2\sqrt{3}}=1+\sqrt{3},$$so $(a,b)=\boxed{(1,3)}.$