7. (6 points) The number in the parentheses that satisfies the equation below is $\qquad$
$$
\frac{\frac{(}{128}+1 \frac{2}{7}}{5-4 \frac{2}{21} \times 0.75} \div \frac{\frac{1}{3}+\frac{5}{7} \times 1.4}{\left(4-2 \frac{2}{3}\right) \times 3}=4.5
$$
【Solution】Solution: Left side $=\frac{\frac{()}{128}+1 \frac{2}{7}}{5-4 \frac{2}{21} \times 0.75} \div \frac{\frac{1}{3}+\frac{5}{7} \times 1.4}{\left(4-2 \frac{2}{3}\right) \times 3}$,
$$
\begin{array}{l}
=\frac{\frac{()}{128}+1 \frac{2}{7}}{5-\frac{43}{14}} \div \frac{\frac{1}{3}+1}{4}, \\
=\frac{\frac{()}{128}+1 \frac{2}{7}}{\frac{27}{14}} \div \frac{1}{3}, \\
=\left[\frac{()}{128}+\frac{9}{7}\right] \times \frac{14}{27} \times 3, \\
=\left[\frac{()}{128}+\frac{9}{7}\right] \times \frac{14}{9}, \\
=\frac{(\quad)}{128} \times \frac{14}{9}+2 .
\end{array}
$$

Therefore, $\frac{(}{128} \times \frac{14}{9}+2=4.5$
Solving for ( $\quad$ ) $=\frac{1440}{7}$.
Thus, the answer is: $\frac{1440}{7}$.