连接$AB$，$CM$，如图所示：

$\because \angle AOB=90^{\circ}$，
$\therefore AB$为$\odot M$的直径，
$\because$点$A\left(8,0\right)$，$B\left(0,6\right)$，
$\therefore OA=8$，$OB=6$，
$\therefore AB=\sqrt{OA^{2}+OB^{2}}=10$，
$\therefore OA=\frac{1}{2}AB=5$，
即$\odot M$的半径为$5$，
$\because M$为圆心，$C$是$\widehat {AO}$的中点，
$\therefore CM\bot OA$，$AD=OD=\frac{1}{2}OA=4$，
$\therefore \angle ADM=90^{\circ}$，
$\therefore DM=\sqrt{A{M}^{2}-A{D}^{2}}=3$，
$\therefore CD=2$，
$\therefore AC=\sqrt{CD^{2}+AD^{2}}=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$，
$\because C$是$\widehat {AO}$的中点，
$\therefore \widehat {AC}=\widehat {OC}$，
$\therefore OC=AC=2\sqrt{5}$，
$\therefore \triangle AOC$的周长为$AO+OC+AC=8+2\sqrt{5}+2\sqrt{5}=4\sqrt{5}+8$.
故答案为：$5$；$4\sqrt{5}+8$.

连接$AB$，$CM$，如图所示：

$\because \angle AOB=90^{\circ}$，
$\therefore AB$为$\odot M$的直径，
$\because$点$A\left(8,0\right)$，$B\left(0,6\right)$，
$\therefore OA=8$，$OB=6$，
$\therefore AB=\sqrt{OA^{2}+OB^{2}}=10$，
$\therefore OA=\frac{1}{2}AB=5$，
即$\odot M$的半径为$5$，
$\because M$为圆心，$C$是$\widehat {AO}$的中点，
$\therefore CM\bot OA$，$AD=OD=\frac{1}{2}OA=4$，
$\therefore \angle ADM=90^{\circ}$，
$\therefore DM=\sqrt{A{M}^{2}-A{D}^{2}}=3$，
$\therefore CD=2$，
$\therefore AC=\sqrt{CD^{2}+AD^{2}}=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$，
$\because C$是$\widehat {AO}$的中点，
$\therefore \widehat {AC}=\widehat {OC}$，
$\therefore OC=AC=2\sqrt{5}$，
$\therefore \triangle AOC$的周长为$AO+OC+AC=8+2\sqrt{5}+2\sqrt{5}=4\sqrt{5}+8$.
故答案为：$5$；$4\sqrt{5}+8$.