6. Given that a sphere is circumscribed around the tetrahedron $ABCD$; another sphere with radius 1 is tangent to the plane $ABC$ and the two spheres are internally tangent at point $D$. Given $AD=3, \cos \angle BAC=$ $\frac{4}{5}, \cos \angle BAD=\cos \angle CAD=\frac{1}{\sqrt{2}}$. Then the volume of the tetrahedron $ABCD$ is
$6 . \frac{18}{5}$.
As shown in Figure 2, first prove that the height $D H$ of the tetrahedron $A B C D$ is a diameter of another sphere. Let $D E \perp A B$, $D F \perp A C$, with the feet of the perpendiculars being $E$ and $F$, respectively, then $A E=A F=A D$. $\cos \angle B A D=\frac{3}{\sqrt{2}}$, $\cos \angle H A E=\cos \angle H A F$ $=\sqrt{\frac{1+\cos \angle B A C}{2}}=$ $\frac{3}{\sqrt{10}}$. Therefore, $A H=\frac{A E}{\cos \angle H A E}=\sqrt{5}, D H=$ $\sqrt{A D^{2}-A H^{2}}=2$.

Thus, the center of the circumscribed sphere of the tetrahedron lies on $D H$, so $A D=B D=C D, A C=A B=2 A E=3 \sqrt{2}$. Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \angle B A C=\frac{27}{5}$.
Hence, $V_{D-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot D H=\frac{18}{5}$