In $\triangle A B C$, the sides $a, b, c$ are opposite to the angles $\angle A, \angle B, \angle C$ respectively. If
$$
2 \sin ^{2} B+3 \sin ^{2} C=2 \sin A \cdot \sin B \cdot \sin C+\sin ^{2} A \text {, }
$$

then $\tan A=$ $\qquad$
5. -1 .

From the Law of Sines, we have
$$
2 b^{2}+3 c^{2}=2 b c \sin A+a^{2} \text {. }
$$

From the Law of Cosines, we have
$$
\begin{array}{l}
2 b^{2}+3 c^{2}=2 b c \sin A+b^{2}+c^{2}-2 b c \cos A \\
\Rightarrow \frac{b^{2}+2 c^{2}}{2 b c}=\sin A-\cos A .
\end{array}
$$

Since $\frac{b^{2}+2 c^{2}}{2 b c} \geqslant \frac{2 \sqrt{b^{2} \cdot 2 c^{2}}}{2 b c}=\sqrt{2}$,
$$
\sin A-\cos A \leqslant \sqrt{2},
$$

Therefore, both equalities hold, i.e., $b=\sqrt{2} c, \angle A=\frac{3 \pi}{4}$.
Thus, $\tan A=-1$.