解:$ (2)$∵$a=7-2\sqrt {6},b=7+2\sqrt {6},$
∴$a+b=14,ab=25.$
∴$a^2+b^2-ab=(a+b)^2-3ab=14^2-3×25$
$=196-75=121.$
$ (3)$∵$\sqrt {16}<2\sqrt {6}=\sqrt {24}<\sqrt {25},$
∴$4<2\sqrt {6}<5,$
∴$-5<-2\sqrt {6}<-4,$
∴$2<7-2\sqrt {6}<3,11<7+2\sqrt {6}<12.$
∵$m $为$a$的整数部分,$n$为$b$的小数部分,
∴$m=2,n=7+2\sqrt {6}-11=2\sqrt {6}-4,$
∴$\frac {m}{n}=\frac {2}{2\sqrt {6}-4}=\frac {\sqrt {6}+2}{2}.$