7. In tetrahedron $ABCD$, it is known that $AB=AC=3, BD=BC=4, BD \perp$ plane $ABC$.

Then the radius of the circumscribed sphere of tetrahedron $ABCD$ is . $\qquad$
7. $\frac{\sqrt{805}}{10}$.

Let the circumcenter of $\triangle ABC$ be $O_{1}$, and the radius be $r$. Let the circumcenter of the tetrahedron $ABCD$ be $O$, and the radius be $R$.
Obviously, $O_{1} O \perp$ plane $ABC$.
As shown in Figure 2, let the midpoint of $BD$ be $E$.

Since $O B=O D$, we have $O E \perp B D$.
Also, $B D \perp$ plane $ABC$, so $O E \parallel$ plane $ABC$.
Thus, $O_{1} O=E B=\frac{1}{2} B D=2$.
In $\triangle ABC$, by the cosine rule we get
$\cos A=\frac{1}{9} \Rightarrow \sin A=\frac{4 \sqrt{5}}{9}$.
By the sine rule we get
$4=2 r \times \frac{4 \sqrt{5}}{9} \Rightarrow r=\frac{9 \sqrt{5}}{10}$.
Therefore, the radius of the circumsphere of the tetrahedron $ABCD$ is
$$
R=\sqrt{r^{2}+2^{2}}=\frac{\sqrt{805}}{10} .
$$