(3) Given that the side length of the base of a regular tetrahedron is 6, and the side edge is 4, then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$ .
(3) 4 Hint: According to the problem, the distance from vertex $A$ to the base $B C D$ is 2. Taking the center of $\triangle B C D$ as $O$, the center of the sphere $P$ lies on the line $A O$. Let the length of $O P$ be $x$, then
$$
x^{2}+12=P B^{2}=P A^{2}=(2-x)^{2},
$$

Solving it yields: $x=-2$. Therefore, the radius of the sphere is 4.