7. In $\triangle A B C$, $A B+A C=7$, and the area of the triangle is 4. Then the minimum value of $\sin A$ is $\qquad$

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7. $\frac{32}{49}$.

From $A B + A C = 7 \Rightarrow A B \cdot A C \leqslant \frac{49}{4}$,
From $\frac{1}{2} A B \cdot A C \sin A = 4 \Rightarrow \sin A \geqslant \frac{32}{49}$.
When $A B = A C = \frac{7}{2}$, the equality holds in both expressions.
Therefore, the minimum value of $\sin A$ is $\frac{32}{49}$.