3. Given the radii of $\odot O_{1}$ and $\odot O_{2}$ are 3 and 5, respectively, and $O_{1} O_{2}=10$. Then the area of the triangle formed by the two inner common tangents and one outer common tangent of the two circles is
3. $\frac{45}{4}$.

As shown in Figure 5, let the internal common tangent $AB$ and $EF$ intersect at $O$, and extend $AB$ and $EF$ to intersect the external common tangent $CD$ at $N$ and $M$. Connect $O_{1}A$ and $O_{2}B$. From $\mathrm{Rt} \triangle A O_{1} O \sim \mathrm{Rt} \triangle \mathrm{BO}_{2} \mathrm{O}$, we get
$$
\left\{\begin{array}{l}
\frac{O_{1} O}{O_{2} O}=\frac{O_{1} A}{O_{2} B}=\frac{3}{5}, \\
O_{1} O+O_{2} O=10 .
\end{array}\right.
$$

Thus, $O_{1} O=\frac{15}{4}, O_{2} O=\frac{25}{4}$.
Therefore, $O A=\sqrt{O_{1} O^{2}-O_{1} A^{2}}=\frac{9}{4}$,
$$
\begin{array}{l}
O B=\sqrt{\mathrm{O}_{2} \mathrm{O}^{2}-\mathrm{O}_{2} \mathrm{~B}^{2}}=\frac{15}{4}, \\
A B=O A+O B=6 .
\end{array}
$$

Connect $\mathrm{O}_{1} \mathrm{C}$ and $\mathrm{O}_{2} \mathrm{D}$, and draw $\mathrm{OG} / / \mathrm{O}_{1} \mathrm{O}_{2}$, then
$$
C D=\sqrt{C G^{2}-G D^{2}}=\sqrt{10^{2}-2^{2}}=4 \sqrt{6} \text {. }
$$

From $A B+B N=N C=C D-N D, B N=N D$, we get
$$
N D=\frac{1}{2}(C D-A B)=2 \sqrt{6}-3 \text {. }
$$

Thus, $M N=C D-2 N D=6$.
Draw $O H \perp M N$ intersecting $C G$ at $P$, then
$$
\begin{array}{l}
\frac{O H-O_{1} C}{O_{2} D-O_{1} C}=\frac{H P}{G D}=\frac{C P}{C G}=\frac{O_{1} O}{O_{1} O_{2}}=\frac{3}{8} \\
O H=3+\frac{3}{8} \times 2=\frac{15}{4}
\end{array}
$$

Thus, $S_{\triangle O M N}=\frac{1}{2} \cdot M N \cdot O H=\frac{1}{2} \times 6 \times \frac{15}{4}=\frac{45}{4}$.