$解:(2)∵3\lt \sqrt {11}\lt 4,2\lt \sqrt {7}\lt 3$
$∴a=\sqrt {11}-3,b=2$
$∴原式=|\sqrt {11}-3-2|+\sqrt {11}=5-\sqrt {11}+\sqrt {11}=5 $
$(3)∵2\lt \sqrt {5}\lt 3$
$∴11\lt 9+\sqrt {5}\lt 12$
$∴x=11,y=\sqrt {5}-2$
$∴x-y=11-\sqrt {5}+2=13-\sqrt {5}$
$∴x-y相反數(shù)為\sqrt {5}-13$
