5. Through the vertex $D$ of the tetrahedron $ABCD$, construct a sphere with radius 1. This sphere is tangent to the circumsphere of the tetrahedron $ABCD$ at point $D$, and is also tangent to the plane $ABC$. If $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$, then the radius of the circumsphere of the tetrahedron $ABCD$ is
5.3.

Draw a perpendicular from point $D$ to plane $A B C$, with the foot of the perpendicular being $H$. Draw $D E \perp A B$ at point $E$, and $D F \perp A C$ at point $F$.
$$
\begin{array}{l}
\text { Then } H E \perp A B, H F \perp A C, \text { and } \\
A E=A F=A D \cos 45^{\circ}=\sqrt{6} . \\
\text { By } \triangle A E H \cong \triangle A F H \Rightarrow \angle H A E=30^{\circ} \\
\Rightarrow A H=\frac{A E}{\cos 30^{\circ}}=2 \sqrt{2}, \\
D H=\sqrt{A D^{2}-A H^{2}}=2 .
\end{array}
$$

Then $D H$ is the diameter of a sphere with radius 1, and the center of the circumscribed sphere of tetrahedron $A B C D$ lies on the extension of $D H$. Let the radius be $r$.
Thus $r^{2}=(r-2)^{2}+(2 \sqrt{2})^{2} \Rightarrow r=3$.