Sphere inside the tetrahedron!

Join the sphere's center to each of the 4 vertices of the tetrahedron. This creates $4$ congruent pyramids. The volume of each pyramid is $\frac{1}{4}$ the volume of the tetrahedron, each having a face in common with the tetrahedron. So, each pyramid has an altitude to that face whose length is $\frac{1}{4}$ that of an altitude of the tetrahedron. This altitude in each pyramid is a radius of the inscribed sphere, therefore its length is $\frac{36}{4} = 9$

A tetrahedron with face area $F$ and altitude (to that face) $a$ has volume $\dfrac{1}{3}aF$

Now, consider a sub-tetrahedron determined by a given face, with fourth vertex at the center of the inscribed sphere. Because the sphere is tangent to the face, the altitude to that face is a radius, say, of length $r$ . If the big tetrahedron is regular (with face area $F$ and altitude $a$ ), then the four sub-tetrahedra are congruent, with volume $\dfrac{1}{3}rF$ .

Together, the sub-tetrahedra fill the big tetrahedron, so the four sub-volumes add up to the big volume:

$4\times \dfrac{1}{3}rF=\dfrac{1}{3}aF$

so that $4r=a$ . You're given that $a=36$ ; consequently, $r=9$ .