2. If four spheres of radius 1 are pairwise tangent, then the radius of the sphere that is tangent to all four spheres is $\qquad$ .
2. $\frac{\sqrt{6}}{2}+1$ or $\frac{\sqrt{6}}{2}-1$.

Since the centers of these four spheres form a regular tetrahedron with an edge length of 2, the distance from the center of the tetrahedron to a vertex is $2 \times \frac{\sqrt{6}}{4}=\frac{\sqrt{6}}{2}$. Therefore, the radius of the circumscribed sphere is $\frac{\sqrt{6}}{2}+1$, and the radius of the inscribed sphere is $\frac{\sqrt{6}}{2}-1$.