$解:(2)∵ \sqrt{9}< \sqrt{13}< \sqrt{16},$
$∴3< \sqrt{13}<4.$
$∴−4<− \sqrt{13}<−3.$
$∴8<5+ \sqrt{13}<9,1<5− \sqrt{13}<2.$
$∴[5+ \sqrt{13}]−8,[5− \sqrt{13}]−1.$
$∴a=5+ \sqrt{13}−8= \sqrt{13} ,b=5− \sqrt{13}−1=4− \sqrt{13}$
$∴a+b= \sqrt{13}−3+4− \sqrt{13}=1$