The given expression is

\[ (8\,{\color{blue}{\div\, 7}}) \times (3\,{\color{red}{\div\, 4}}). \]

Since our expression contains multiplication and division only, we can write an equivalent expression by swapping the ${\color{red}{\div\, 4}}$ and the ${\color{blue}{\div\, 7}}$:

\[ (8\, {\color{red}{\div\, 4}}) \times(3\, {\color{blue}{\div\, 7}}) \]

We can now interpret the divisions as fractions:

\[ \dfrac{8} 4\times \dfrac 3 {7} \]

So the missing fraction is $ \dfrac{8} 4.$

The given expression is

\[ (3\,{\color{red}{\div\, 2}})\times (4\,{\color{blue}{\div\, 5}}). \]

Since our expression contains multiplication and division only, we can write an equivalent expression by swapping the ${\color{red}{\div\, 2}}$ and the ${\color{blue}{\div\, 5}}$:

\[ (3\, {\color{blue}{\div\, 5}}) \times (4\, {\color{red}{\div\, 2}}) \]

We can now interpret the divisions as fractions:

\[ \dfrac 3 5 \times \dfrac 4 2 \]

So the missing fraction is $\dfrac 4 2.$

Interpreting the fraction as division, we can write our expression as

\[ (5\,{\color{red}{\div\, 6}}) \times (12\,{\color{blue}{\div\, 5}}). \]

We use parentheses to make it easier to read.

Since our expression contains multiplication and division only, we can write an equivalent expression by swapping the ${\color{red}{\div\, 6}}$ and the ${\color{blue}{\div\,5}}$:

\[ (5\,{\color{blue}{\div\,5}}) \times (12\,{\color{red}{\div\, 6}}) \]

Finally, we can write the $5\div 5$ and $12\div 6$ as fractions:

\[ \dfrac{5}{5} \times \dfrac{12}{6} \]

Interpreting the fraction as division, we can write our expression as

\[ (6\,{\color{red}{\div\, 5}}) \times (10\,{\color{blue}{\div\, 3}}). \]

We use parentheses to make it easier to read.

Since our expression contains multiplication and division only, we can write an equivalent expression by swapping the ${\color{red}{\div\, 5}}$ and the ${\color{blue}{\div\,3}}$:

\[ (6\,{\color{blue}{\div\,3}}) \times (10\,{\color{red}{\div\, 5}}) \]

Finally, we can write the $6\div 3$ and $10\div 5$ as fractions:

\[ \dfrac{6}{3} \times \dfrac{10}{5} \]

Interpreting the fractions as division, we get

\[ (27\,{\color{red}{\div\, 4}})\times(8\,{\color{blue}{\div\, 9}}). \]

Since our expression contains multiplication and division only, we can create an equivalent expression by swapping the ${\color{red}{\div\, 4}}$ and ${\color{blue}{\div\, 9}}$:

\[ (27\,{\color{blue}{\div\, 9}})\times(8\,{\color{red}{\div\, 4}}) \]

Interpreting the divisions as fractions, we get

\[ \dfrac{27}{9}\times\dfrac{8}{4}. \]

Carrying out the divisions, we arrive at

\begin{align*} \underbrace{\dfrac{27}{9}}_{3}\times\underbrace{\dfrac{8}{4}}_{2} = 3\times 2 = 6. \end{align*}

Interpreting the fractions as division, we get

\[ (12\,{\color{red}{\div\, 7}})\times(35\,{\color{blue}{\div\, 6}}). \]

Since our expression contains multiplication and division only, we can create an equivalent expression by swapping the ${\color{red}{\div\, 7}}$ and ${\color{blue}{\div\, 6}}$:

\[ (12\,{\color{blue}{\div\, 6}})\times(35\,{\color{red}{\div\, 7}}) \]

Interpreting the divisions as fractions, we get

\[ \dfrac{12}{6}\times\dfrac{35}{7}. \]

Carrying out the divisions, we arrive at

\begin{align*} \underbrace{\dfrac{12}{6}}_{2}\times\underbrace{\dfrac{35}{7}}_{5} = 2\times 5 = 10. \end{align*}

We need to solve the following multiplication problem:

\[ \dfrac{1}{\color{red}10} \times \dfrac{\color{blue}4}{5} \]

Notice that the first fraction's denominator ($\color{red}10$) and the second fraction's numerator ($\color{blue}4$) have a common factor of $2.$

Therefore, we can simplify our problem by first swapping the denominators:

\[ \dfrac{1}{5} \times \dfrac{\color{blue}4}{\color{red}10} \]

Next, we reduce the second fraction to its lowest terms by dividing the numerator and denominator by $2.$

\begin{align*} \dfrac15 \times \dfrac{{\color{blue}{4}}\div 2}{{\color{red}{10}}\div 2} &=\dfrac15\times \dfrac25 \end{align*}

To multiply two fractions, we multiply the numerators, and we multiply the denominators:

\[ \dfrac{1}{5} \times \dfrac{2}{5} = \dfrac{1 \times 2}{5 \times 5} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{2}{25}} \]

We need to solve the following multiplication problem:

\[ \dfrac{5}{\color{red}9} \times \dfrac{\color{blue}6}{7} \]

Notice that the first fraction's denominator ($\color{red}9$) and the second fraction's numerator ($\color{blue}6$) have a common factor of $3.$

Therefore, we can simplify our problem by first swapping the denominators:

\[ \dfrac{5}{7} \times \dfrac{\color{blue}6}{\color{red}9} \]

Next, we reduce the second fraction to its lowest terms by dividing the numerator and denominator by $3.$

\begin{align*} \dfrac57 \times \dfrac{{\color{blue}{6}}\div 3}{{\color{red}{9}}\div 3} &=\dfrac57\times \dfrac23 \end{align*}

To multiply two fractions, we multiply the numerators, and we multiply the denominators:

\[ \dfrac{5}{7} \times \dfrac{2}{3} = \dfrac{5 \times 2}{7 \times 3} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{10}{21}} \]

We need to solve the following multiplication problem:

\[ \dfrac{4}{\color{red}81} \times \dfrac{\color{blue}36}{7} \]

Notice that the first fraction's denominator ($\color{red}81$) and the second fraction's numerator ($\color{blue}36$) have a common factor of $9.$

Therefore, we can simplify our problem by first swapping the denominators:

\[ \dfrac{4}{7} \times \dfrac{\color{blue}36}{\color{red}81} \]

Next, we reduce the second fraction to its lowest terms by dividing the numerator and denominator by $9.$

\begin{align*} \dfrac{4}{7} \times \dfrac{{\color{blue}{36}}\div 9}{{\color{red}{81}}\div 9} &=\dfrac{4}{7}\times \dfrac49 \end{align*}

To multiply two fractions, we multiply the numerators, and we multiply the denominators:

\[ \dfrac{4}{7} \times \dfrac{4}{9} = \dfrac{4 \times 4}{7\times 9} = \bbox[3pt,Gainsboro]{\color{blue}\dfrac{16}{63}} \]