(6) \( \begin{aligned} {[3 \div(2-1)]+\frac{5}{7} } & =[\square \div \square]+\frac{5}{7} \\ & =\square+\frac{5}{7} \\ & =3 \frac{5}{7}\end{aligned} \)

To solve for the values represented by the squares in the equation, let's break it down. The first part, \( 3 \div (2-1) \), simplifies to \( 3 \div 1 = 3 \). Therefore, the first square can be found as follows: since this equals \( [\square \div \square] \), we can set it equal to 3. Next, we know that \( \square + \frac{5}{7} = 3 \frac{5}{7} \). Converting \( 3 \frac{5}{7} \) into an improper fraction gives us \( \frac{26}{7} \). Now, rearranging to find the other square, we have \( \square = \frac{26}{7} - \frac{5}{7} = \frac{21}{7} = 3 \). So we can fill in the blanks: \[ \begin{aligned} \text{The first square: } & 3 \\ \text{The second square: } & 3 \end{aligned} \] The complete equation now reads \( [3 \div (2-1)] + \frac{5}{7} = [3 \div 1] + \frac{5}{7} = 3 + \frac{5}{7} = 3 \frac{5}{7} \).