Consider given rational number $\frac{4}{7}$.
To make the denominator as 63, we have to multiply it with 9.

Multiplying both numerator and denominator by 9, we get

$\frac{4\times 9}{7\times 9}=\frac{36}{63}$
Therefore, $\frac{36}{63}$ is an equivalent rational number of  $\frac{4}{7}$ with denominator equal to 63.

$\text{Reciprocal of}\frac{-2}{3}\text{is}\frac{-3}{2}.$

$\frac{2}{3}-\frac{3}{2}=\frac{4}{6}-\frac{9}{6}=\frac{(4-9)}{6}=\frac{-5}{6}$
 [LCM of 2 and 3 is 6]

$\frac{13}{14}$
is less than 1 as the numerator of the fraction is smaller than the denominator.
While on the other hand, the other fractions are greater than 1 as all the numbers are positive and their numerator is greater than the denominator.

Any negative number is less than 0. So, $\frac{-4}{3}<0$

12 $\times$ 9 = 108
$\Rightarrow \frac{108}{12}=\frac{(12\times 9)}{12}$

= 9 

Let the number to be added be x.

$\frac{-3}{8}+x=\frac{5}{16}$

$x=\frac{5}{16}+\frac{3}{8}=\frac{5}{16}+\frac{6}{16}=\frac{11}{16}$

Reducing  $\frac{39}{26}$   to its lowest form ,  we get, 

    $\frac{39÷13}{26÷13}$ = $\frac{3}{2}$( 13 is the common factor)
Multiplying numerator and denominator by 2 we get
$\frac{3\times 2}{2\times 2}$ = $\frac{6}{4}$
Hence, $\frac{39}{26}$ is also equal to  $\frac{6}{4}$
So, both the fraction are equal
Hence,  $\frac{39}{26}=\frac{6}{4}$
                   

The space between 0 and 1 is divided into 10 equal parts. The ball lies in sixth place.

Therefore, the location ball lies in is, $\frac{6}{10}=\frac{3}{5}$

$-\frac{16}{3}÷-\frac{4}{3}=-\frac{16}{3}\times -\frac{3}{4}=4$

Consider the given rational number $\frac{4}{7}$.
To make the numerator as 16, we have to multiply it by 4.

Multiplying both numerator and denominator by 4, we get

$\frac{4\times 4}{7\times 4}=\frac{16}{28}$
Therefore, $\frac{16}{28}$ is an equivalent rational number of $\frac{4}{7}$ with numerator equal to 16.