Answer: 90

Small Hint:

Let $F$ be the foot of the perpendicular from $L$ to $\overline{KN};$ in right triangle $KLN$ the altitude gives $KF \cdot KN = KL^2$

Big Hint:

Triangles $KFO$ and $KMN$ share angle $K$ and each has a right angle, so $KF \cdot KN = KO \cdot KM$

Solution:

Let $F$ be the foot of the perpendicular from $L$ to $\overline{KN},$ so $O$ lies on segment $LF.$ In right triangle $KLN$ (right angle at $L$), the altitude $LF$ to the hypotenuse gives the geometric mean relation $KF \cdot KN = KL^2 = 28^2 = 784.$

Triangles $KFO$ and $KMN$ share angle $K,$ and $\angle KFO = 90^\circ = \angle KMN,$ so they are similar. Hence $\frac{KF}{KM} = \frac{KO}{KN},$ that is, $KO \cdot KM = KF \cdot KN = 784.$ With $KO = 8$ this gives $KM = 98,$ so $$\begin{aligned} MO &= KM - KO \\ &= 98 - 8 = 90. \end{aligned}$$