12. In the tetrahedron $ABCD$, it is known that $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. The radius of the sphere that passes through $D$ and is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron is 1, then the radius of the circumscribed sphere of the tetrahedron $ABCD$ is
12.3.

As shown in Figure 3, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ and $DF \perp AC$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $HE \perp AB$, $HF \perp AC$, and
$AE = AF = AD \cos 45^{\circ} = \sqrt{6}$.
From $\triangle AEH \cong \triangle AFH \Rightarrow \angle HAE = 30^{\circ}$.
Thus, $AH = \frac{AE}{\cos 30^{\circ}} = 2\sqrt{2}$,
$$
DH = \sqrt{AD^2 - AH^2} = 2,
$$

Therefore, $DH$ is the diameter of a sphere with radius 1.
Thus, the center $O$ of the circumscribed sphere of tetrahedron $ABCD$ lies on the extension of $DH$.
Let the radius of the circumscribed sphere be $r$. Then
$$
r^2 = (r-2)^2 + (2\sqrt{2})^2 \Rightarrow r = 3.
$$