$\begin{align}  & {{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab \\ & a+b=\frac{1}{2-\sqrt{3}}+\frac{1}{2+\sqrt{3}}=\frac{2+\sqrt{3}+2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} \\ & (a+b)=\frac{4}{4-3}=4 \\ & {{(a+b)}^{2}}={{4}^{2}}=16 \\ & 2ab=2\left( \frac{1}{2-\sqrt{3}}\times \frac{1}{2+\sqrt{3}} \right)=2\left( \frac{1}{4-3} \right)=2 \\ & {{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab=16-2=14 \\\end{align}$