7. $\frac{18}{5}$.

First, prove that the height $D H$ of tetrahedron $A B C D$ is the diameter of the small sphere.

Draw $H E \perp A B$ at point $E$, and $H F \perp A C$ at point $F$. Then
$$
\begin{array}{l}
A E=A F=A D \cos \angle D A B=\frac{3}{\sqrt{2}}, \\
\cos \angle H A E=\sqrt{\frac{1+\cos \angle B A C}{2}}=\frac{3}{\sqrt{10}} \\
\Rightarrow A H=\frac{A E}{\cos \angle H A E}=\sqrt{5} \\
\Rightarrow D H=\sqrt{A D^{2}-A H^{2}}=2 .
\end{array}
$$

Thus, the center of the large sphere is also on $D H$.
Therefore, $A D=B D=C D, A B=A C=3 \sqrt{2}$.
So, $S_{\triangle A B C}=\frac{27}{5}, V_{\text {tetrahedron } A B C D}=\frac{18}{5}$.
7. A sphere is circumscribed around the tetrahedron $ABCD$, and another sphere with radius 1 is tangent to the plane $ABC$ and the two spheres are internally tangent at point $D$. If
$$
\begin{array}{l}
AD=3, \cos \angle BAC=\frac{4}{5}, \\
\cos \angle BAD=\cos \angle CAD=\frac{1}{\sqrt{2}},
\end{array}
$$

then the volume of the tetrahedron $ABCD$ is