$解:∵\frac{1}{3-\sqrt {7} }=\frac{3+7}{(3+\sqrt{7})(3-\sqrt {7} )} =\frac{3+\sqrt {7} }{2},$
$又2<\sqrt {7}< 3,∴\frac{5}{2}<\frac{3+\sqrt {7} }{2}<3,\ $
$∴a=2,b=\frac{3+\sqrt{7}}{2}-2=\frac{\sqrt {7} -1}{2},$
$∴\frac{a}{b}=2÷\frac {\sqrt {7} -1}{2}=\frac{4}{\sqrt {7} -1}=\frac{2\sqrt {7} +2}{3}.$