7. In $\triangle A B C$, the side lengths opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$, respectively, and
$$
\begin{array}{l}
\sin C \cdot \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}, \\
\cos A=\frac{3}{5}, a=4 .
\end{array}
$$

Then the area of $\triangle A B C$ is . $\qquad$
7.6 .

From equation (1) we know
$$
\begin{array}{l}
2 \sin \frac{A}{2}=\sin \left(C+\frac{A}{2}\right) \\
\Rightarrow 2 \sin A=2 \sin \left(C+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$

Thus, $c+b=2a$.
Also, $a^{2}=b^{2}+c^{2}-2 b c \cos A$, which means
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-b) \frac{3}{5}
$$
$\Rightarrow b=3$ or 5, corresponding to $c=5$ or 3.
Therefore, $S_{\triangle A B C}=\frac{1}{2} b c \sin A$
$$
=\frac{1}{2} \times 3 \times 5 \times \frac{4}{5}=6 \text {. }
$$