$解:(1)当a=\sqrt 3+ \sqrt{2},b=\sqrt 3-\sqrt{2} 时, $
$ \begin{aligned} 原式&=( \sqrt{3}+\sqrt 2)-(\sqrt 3-\sqrt 2) \\ &=\sqrt{3}+\sqrt 2-\sqrt 3+\sqrt 2 \\ &=2 \sqrt{2}. \\ \end{aligned}$
$(2)当a=\sqrt{3}+\sqrt 2,b=\sqrt{3}-\sqrt 2时, $
$ \begin{aligned} 原式&=(a-b)^2-ab \\ &=(2\sqrt{2})^2-(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) \\ &=8-(3-2) \\ &=8-1 \\ &=7. \\ \end{aligned}$