(4) As shown in the figure, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, the volume $V$ of the tetrahedron $D-ABC$ is $\qquad$
(4) 2 Hint: From $D A \perp$ plane $A B C$, we know that $D A \perp A B$, and plane $D A B \perp$ plane $A B C$.

As shown in the figure, take the midpoint $O$ of $A B$. Then, from the fact that $\triangle A B C$ is an equilateral triangle with side length 2, we know that $C O \perp A B, C O \perp$ plane $D A B$, and $C O=\sqrt{3}$.
Construct $O E \perp D B$ at $E$, and connect $C E$, then $C E \perp D B$. Therefore, $\angle O E C$ is the plane angle of the dihedral angle $A-B D-C$. Let $D A=x$, then
$$
O E=\frac{1}{2} \times \frac{D A \times A B}{D B}=\frac{1}{2} \times \frac{2 x}{\sqrt{x^{2}+4}}=\frac{x}{\sqrt{x^{2}+4}},
$$

From $\tan \angle O E C=\frac{O C}{O E}=2$, we know that $\sqrt{3} \times \sqrt{x^{2}+4}=2 x$, solving for $x$ gives $x=2 \sqrt{3}$. Therefore, the volume of the tetrahedron $D-A B C$ is $V=\frac{1}{3} \times 2 \sqrt{3} \times\left(\frac{\sqrt{3}}{4} \times 2^{2}\right)=2$.