解:
(1) $\because$ 在$△ ABC$中,$∠ ACB=90°,$
$\therefore AC^2 + BC^2 = AB^2。$
$\because AC=5,$$BC=12,$
$\therefore AB^2 = 5^2 + 12^2 = 169,$
$\therefore AB=13。$
(2) $\because ∠ ACB=90°,$$CO⊥ AB,$
$\therefore S_{△ ABC} = \frac{1}{2}AC· BC = \frac{1}{2}AB· CO,$
即$AC· BC = AB· CO,$
$\therefore 5×12 = 13CO,$
$\therefore CO=\frac{60}{13}。$
$\because$ 在$\mathrm{Rt}△ AOC$中,$AO^2 + CO^2 = AC^2,$
$\therefore AO^2 = AC^2 - CO^2 = 5^2 - (\frac{60}{13})^2 = \frac{625}{169},$
$\therefore AO=\frac{25}{13}。$