7. Let $z=x+y \mathrm{i}(x, y \in \mathbf{R}, \mathrm{i}$ be the imaginary unit), the imaginary part of $z$ and the real part of $\frac{z-\mathrm{i}}{1-z}$ are both non-negative. Then the area of the region formed by the points $(x, y)$ on the complex plane that satisfy the conditions is

 untranslated part:
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Note: The last part is a note and not part of the problem statement, so it is not translated.
7. $\frac{3 \pi+2}{8}$.

Notice that,
$$
\begin{array}{l}
\operatorname{Re} \frac{z-\mathrm{i}}{1-z}=\operatorname{Re} \frac{x+(y-1) \mathrm{i}}{1-x-y \mathrm{i}} \\
=\frac{x(1-x)-(y-1) y}{(1-x)^{2}+y^{2}} \geqslant 0 \\
\Leftrightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2} .
\end{array}
$$

Thus, the set of points satisfying the condition is the disk
$$
\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}
$$

in the upper half of the real axis. The area of this part is calculated to be $\frac{3 \pi+2}{8}$.