5. A regular triangle with a side length of 2 is divided into four smaller triangles by its median lines. Except for the central small triangle, three other small triangles are used as bases to construct three regular tetrahedrons of height 2 on the same side. There is a sphere that is tangent to the plane of the original triangle and also tangent to one side face of each tetrahedron. What is the radius of this small sphere?
Let the center of the sphere be point $O$, and the radius be $R$. By symmetry, the point where the sphere $O$ touches the base $\triangle ABC$ is the centroid of $\triangle ABC$.

For the regular tetrahedron $O_{1}-AEF$, the foot of the perpendicular from the apex to the base, $M$, is the centroid of $\triangle AEF$. Points $M$ and $I$ lie on segment $AD$, which intersects $EF$ at point $P$ (refer to the figures above).
It is evident that $AM = 2MP = 2PI = ID$.
Let the sphere $O$ touch the tetrahedron $O_{1}-AEF$ at point $Q$, then
$$
OQ = OI = R.
$$

Given $AB = 2$, we have $AD = \sqrt{3}$. Thus,
$$
MP = PI = \frac{\sqrt{3}}{G_{0}}.
$$

Let $\angle MPO_{1} = \alpha$, $\angle OPI = \beta$, and $\alpha + 2\beta = \pi$, then $\lg \alpha = \frac{2}{\sqrt{3}} = 4\sqrt{3}$,
$$
\begin{array}{l}
\operatorname{tg} 2\beta = \operatorname{tg}(\pi - \alpha) = -4\sqrt{3}. \\
\text{Also, } \operatorname{tg} 2\beta = \frac{2 \operatorname{tg} \beta}{1 - \operatorname{tg}^2 \beta} = \frac{2 \frac{R}{PI}}{1 - \frac{R^2}{RI^2}} = \frac{\sqrt{3} R}{\frac{1}{12} - R^2}
\end{array}
$$

Solving this, we get $R = \frac{1}{3}$.