7. (6 points) In the following (1), (2), (3) equations, each has a $\square$. Please indicate in the $\qquad$th equation, $\square=3 \frac{1}{20}$.
(1) $\left.4 \frac{1}{2} \div\left[0.4+\left(6 \frac{1}{3}-\square\right) \div 12 \frac{1}{2} \times 0.75\right]=10 \div 12 \frac{1}{2} \times 0.75\right]=10$
(2) $\frac{\left[\frac{5}{8} \times\left(2 \frac{3}{4}-1.15\right)+7.8\right] \div 8 \frac{4}{5}}{(6-\square) \div 29.5}$
(3) $\left[\left(6.5-\frac{2}{3}\right) \div 3 \frac{1}{2}-1 \frac{8}{15}\right] \times(\square+71.95)=1$.
【Solution】Solution: (1) $4 \frac{1}{2} \div\left\{0.4+\left(6 \frac{1}{3}-3 \frac{1}{20}\right) \div 12 \frac{1}{2} \times 0.75\right\}$
$$
\begin{array}{l}
=4.5 \div\left\{0.4+3 \frac{17}{60} \div 12 \frac{1}{2} \times 0.75\right\} \\
=4.5 \div\left\{0.4+\frac{197}{1000}\right\} \\
=4.5 \div\{0.4+0.197\} \\
=4.5 \div 0.597 \\
\approx 7.538 ; \\
=\frac{\left.[2) \frac{5}{8} \times\left(2 \frac{3}{4}-1.15\right)+7.8\right] \div 8 \frac{4}{5}}{(6-\square) \div 29.5} \\
=\frac{8.8 \div 8.8}{2.95 \div 29.5} \\
=10 ;
\end{array}
$$
$$
\begin{aligned}
& \text { (3) }\left\{\left(6.5-\frac{2}{3}\right) \div 3 \frac{1}{2}-1 \frac{8}{15}\right\} \times\left(3 \frac{1}{20}+71.95\right) \\
= & \left\{5 \frac{5}{6} \div 3 \frac{1}{2}-1 \frac{8}{15}\right\} \times 75 \\
= & \left\{\frac{5}{3}-1 \frac{8}{15}\right\} \times 75 \\
= & \frac{2}{15} \times 75 \\
= & 10 ;
\end{aligned}
$$

Answer: After substitution, the results of the three expressions are $7.538, 10, 10$.
Therefore, the answer is: (3).