In triangle ABC, by applying pythagoras rule,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=8^2+6^2=64+36$
$AC=\sqrt{64+36}=\sqrt{100}=10cm$
Since, M is the mid-point of AC,
$AM=MC=\frac{AC}{2}=\frac{10}{2}=5cm$
We know that in a right angled triangle, mid-point of the hypotenuse is always at equidistance from all the three vertices of the triangle.
 $\Rightarrow$ AM = MC = MB = 5 cm