Consider one diagonal of the isosceles trapezoid.

Its horizontal projection is \(\dfrac{12+6}{2}=9\text{ cm}\), and the height is \(3\sqrt{7}\text{ cm}\).

So the diagonal length is:

\(d=\sqrt{9^2+(3\sqrt{7})^2}=\sqrt{81+63}=12\text{ cm}\).

The intersection point of the diagonals divides them in the ratio of the bases:

\(12:6=2:1\).

Therefore, a diagonal of length \(12\text{ cm}\) is divided into segments \(8\text{ cm}\) and \(4\text{ cm}\).

The hypotenuse is \(\sqrt{9^2+12^2}=15\text{ cm}\).

The smaller acute angle is opposite the shorter leg, so point \(M\) divides the hypotenuse in the ratio \(5:10=1:2\).

Place the triangle so that the right-angle vertex is \((0,0)\), and the ends of the legs are \((12,0)\) and \((0,9)\).

Then the point \(M\), which is \(5\text{ cm}\) from the smaller-angle vertex, has coordinates \((8,3)\).

Hence the distance to the right-angle vertex is:

\(OM=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}\text{ cm}\).

The centers of circles inscribed in the same angle lie on the angle bisector.

The distances from the angle vertex to the centers are proportional to the radii.

Let the distance to the center of the smaller circle be \(x\). Then the distance to the center of the larger circle is \(\dfrac{6}{4}x=\dfrac{3}{2}x\).

Since the circles do not touch each other, the distance between their centers equals the difference of these distances:

\(\dfrac{3}{2}x-x=13\).

\(\dfrac{x}{2}=13\), so \(x=26\text{ cm}\).

Place point \(B\) at the origin and side \(BC\) on the \(x\)-axis.

Then \(B=(0,0)\), \(C=(7,0)\), \(K=(2,0)\).

Find the coordinates of \(A=(x,y)\) from:

\(AB=6\Rightarrow x^2+y^2=36\), \(AC=8\Rightarrow (x-7)^2+y^2=64\).

Subtracting, we get \(x=\dfrac{3}{2}\), and then \(y=\dfrac{3\sqrt{15}}{2}\).

Point \(M\) is the midpoint of \(AB\), so \(M=\left(\dfrac{3}{4},\dfrac{3\sqrt{15}}{4}\right)\).

Then

\(MK^2=\left(2-\dfrac34\right)^2+\left(\dfrac{3\sqrt{15}}{4}\right)^2=\dfrac{25}{16}+\dfrac{135}{16}=10\).

Therefore, \(MK=\sqrt{10}\text{ cm}\).

The centers of the circles lie on the angle bisector, and the distances from the vertex to the centers are proportional to the radii.

Let the distance to the center of the smaller circle be \(x\). Then the distance to the center of the larger circle is \(\dfrac{8}{6}x=\dfrac{4}{3}x\).

Since the circles are externally tangent, the distance between the centers equals the sum of the radii:

\(6+8=14\text{ cm}\).

On the other hand, this distance equals the difference of the distances from the vertex:

\(\dfrac{4}{3}x-x=14\).

\(\dfrac{x}{3}=14\), hence \(x=42\text{ cm}\).