Solution:

Let the tetrahedron have side $1$. Then the center $O$ is a distance $1 / \sqrt{8}$ from the center of each of the spheres, so it is contained in each of the spheres. We now use convexity.

Two circles with diameters two of the sides of a triangle cover the triangle (consider the foot of the altitude to the third side), so faces $ABC$ and $ABD$ are certainly contained in the spheres. Consider face $ACD$. The sphere on $BC$ passes through the midpoints of $AC$ and $CD$, and through $C$, so it contains the triangle formed by these three points (by convexity). But the rest of $ACD$ is contained in the sphere on $AD$. Similarly for the face $BCD$. Hence all the faces are contained in the spheres. But now take any point $P$ inside the tetrahedron. Extend $OP$ to meet a face at $X$. $X$ lies in one of the spheres, but $O$ also lies in the sphere and hence all points on $OX$, including $P$ (by convexity).

False in general. Take $ABCD$ to be a plane square, then no points on $CD$ are in the spheres except $C$ and $D$ (and we can obviously distort this slightly to make it less degenerate).