3. Two spheres $O_{1}$ and $O_{2}$, both with a radius of 1, are tangent to each other, and they are also tangent to the two half-planes of a dihedral angle $\alpha-l-\beta$ of $60^{\circ}$. There is another larger sphere $O$ that is tangent to both half-planes of the dihedral angle and is externally tangent to spheres $O_{1}$ and $O_{2}$. What is the radius $R$ of sphere $O$?
3. $\frac{5+\sqrt{13}}{3}$.

As shown in Figure 8, it is easy to see that the centers of the three spheres $O$, $O_{1}$, and $O_{2}$ are all on the bisector plane of the dihedral angle $\alpha-l-\beta$. Let the sphere $O_{1}$ be tangent to the planes $\alpha$ and $\beta$ at points $A$ and $B$ respectively, and the sphere $O$ be tangent to the planes $\alpha$ and $\beta$ at points $D$ and $E$ respectively. Draw $A C \perp l$ at point $C$, and $D F \perp l$ at point $F$. Then $B C \perp l$, $E F \perp l$, and $\angle O C A = \angle O F D = 30^{\circ}$.
Thus, $O_{1} C = 2$, $O F = 2 R$.
Connect $O_{1} O_{2}$ intersecting $O F$ at point $H$, then $H$ is the point of tangency between sphere $O_{1}$ and sphere $O_{2}$, and
$$
\begin{array}{l}
O H \perp O_{1} O_{2}, O H = 2 R - 2, O_{1} H = 1, \\
O O_{1} = R + 1 .
\end{array}
$$

In the right triangle $\triangle O H O_{1}$, $O H^{2} + O_{1} H^{2} = O O_{1}^{2}$, i.e., $(2 R - 2)^{2} + 1 = (R + 1)^{2}$.
Solving this, we get $R = \frac{5 \pm \sqrt{13}}{3}$.
Since $R > 1$, we have $R = \frac{5 + \sqrt{13}}{3}$.