$解:(1)∵\sqrt {a-1}+\sqrt {b+3}=0,∴a-1=0,b+3=0,$
$∴a=1,b=-3,∴a+b=1+(-3)=-2.$
$(2)∵\sqrt {y-5}≥0,\sqrt{5-y}≥0,$
$∴y≥5且y≤5.∴y=5.$
$由x^2=\sqrt {y-5}+\sqrt{5-y}+9.$
$知x^2=9,于是x=±3.$
$当x=3,y=5时,x+y=8;$
$当x=-3,y=5时,x+y=2.$
$综上,x+y的值为8或2.$
$(3)∵|2m-4|+|n+2|+\sqrt{(m-3)n^2}+4=2m,$
$∴(m-3)n^2≥0.$
$∴m≥3.$
$∴2m-4>0.$
$∴2m-4+|n+2|+\sqrt{(m-3)n^2}+4=2m.$
$∴|n+2|+\sqrt{(m-3)n^2}=0.$
$又∵|n+2|≥0,\sqrt{(m-3)n^2}≥0,$
$∴n+2=0且(m-3)n^2=0.$
$∴m=3,n=-2.$
$∴m+n=3-2=1.$