Given, -2 and -1
Let -2 and -1 be rational numbers with denominator 6.
Then, we have -2 and -1 = \(\\ \frac { -12 }{ 6 } \) and \(\frac { -6 }{ 6 } \)
So,\(\frac { -12 }{ 6 } <\frac { -11 }{ 6 } <\frac { -10 }{ 6 } <\frac { -9 }{ 6 } <\frac { -8 }{ 6 } <\frac { -7 }{ 6 } <\frac { -6 }{ 6 } \) 
or \(\\ -2<\frac { -11 }{ 6 } <\frac { -5 }{ 3 } <\frac { -3 }{ 2 } <\frac { -4 }{ 3 } <\frac { -7 }{ 6 } <-1\)
Hence, the five rational numbers between -2 and -1
are \(\frac { -11 }{ 6 } ,\frac { -5 }{ 3 } ,\frac { -3 }{ 2 } ,\frac { -4 }{ 3 }\) and \(\frac { -7 }{ 6 } \).

We have, \(\frac { -2 }{ 3 } +0\) = \(\frac { -2 }{ 3 } +0=\frac { -2 }{ 3 } +\frac { 0 }{ 3 } =\frac { -2+0 }{ 3 } =\frac { -2 }{ 3 } \)[ we can take any number as denominator of 0]

\(\frac { 5 }{ 3 } +\frac { 3 }{ 5 } \)
Here, the denominator are not same.
\(\because \) LCM of denominator 3 and 5 = 15
\(\therefore \) \(\frac { 5 }{ 3 } =\frac { 5\times 5 }{ 3\times 5 } =\frac { 25 }{ 15 } \)
Which is an equivalent rational number. 
and \(\frac { 3 }{ 5 } =\frac { 3\times 3 }{ 5\times 3 } =\frac { 9 }{ 15 } \) 
Which is an equivalent rational number.
Now, \(\frac { 5 }{ 3 } +\frac { 3 }{ 5 } =\frac { 25 }{ 15 } +\frac { 9 }{ 15 } =\frac { 25+9 }{ 15 } =\frac { 34 }{ 15 } \)

Here, the denominators of given rational numbers are not same.

\(\because \) LCM = 2\(\times \)2\(\times \)7=28
\(\therefore \frac { -3 }{ 7 } =\frac { -3\times 4 }{ 7\times 4 } =\frac { -12 }{ 28 } ,\frac { -3 }{ 2 } \) 
\(=\frac { -3\times 14 }{ 2\times 14 } =\frac { -42 }{ 28 } \)
​​​​​​​and \(\\ \frac { -3 }{ 4 } =\frac { -3\times 7 }{ 4\times 7 } =\frac { -21 }{ 28 } \)
We observe that, 12 < 21 < 42
\(\Rightarrow \frac { 12 }{ 28 } <\frac { 21 }{ 28 } <\frac { 42 }{ 28 } \)
\(\Rightarrow \frac { -12 }{ 28 } >\frac { -21 }{ 28 } >\frac { -42 }{ 28 } \)
\(\Rightarrow \frac { -3 }{ 7 } >\frac { -3 }{ 4 } >\frac { -3 }{ 2 } \)
Hence, the given rational numbers, when arranged in ascending order are \(\frac { -3 }{ 2 } ,\frac { -3 }{ 4 } ,\frac { -3 }{ 7 } \).

Given, \(\frac { -1 }{ 4 } ,\frac { 1 }{ 4 } \)
We know that, every positive rational number is greater than every negative rational number.
Hence, \(\\ \frac { 1 }{ 4 } \) is greater than \(\\ \frac { -1 }{ 4 } \).

Given, 0 \(\boxed { } \)\(\frac { -7 }{ 6 } \)
We know that, a negative rational number is always to the left of zero. So, Zero is always greater than any negative rational number.
Hence , 0 > \(\frac { -7 }{ 6 } \)

\(\frac { 8 }{ -5 } \) and \(\frac { -24 }{ 15 } \) represent the same rational number 

Given,\(\frac { -3 }{ 5 } ,\frac { -6 }{ 10 } ,\frac { -9 }{ 15 } ,\frac { -12 }{ 20 } ,....\)
Here,\(\frac { -3 }{ 5 } \) is the rational number in standard form.
Now, \(\frac { -3 }{ 5 } =\frac { -3\times 1 }{ 5\times 1 } ,\frac { -6 }{ 10 } =\frac { -3\times 2 }{ 5\times 2 } ,\) 
\(\frac { -9 }{ 15 } =\frac { -3\times 3 }{ 5\times 3 } ,\frac { -12 }{ 20 } =\frac { -3\times 4 }{ 5\times 4 } \)
 or \(\frac { -3\times 1 }{ 5\times 1 } =\frac { -3 }{ 5 } ,\frac { -3\times 2 }{ 5\times 2 } =\frac { -6 }{ 10 } \)
\(\frac { -3\times 3 }{ 5\times 3 } =\frac { -9 }{ 15 } ,\frac { -3\times 4 }{ 5\times 4 } =\frac { -12 }{ 20 } \)
Thus, we observe a pattern in these numbers. The next four numbers are
\(\frac { -3\times 5 }{ 5\times 5 } =\frac { -15 }{ 25 } ,\frac { -3\times 6 }{ 5\times 6 } =\frac { -18 }{ 30 } \)
\(\\ \frac { -3\times 7 }{ 5\times 7 } =\frac { -21 }{ 35 } ,\frac { -3\times 8 }{ 5\times 8 } =\frac { -24 }{ 40 } \)
Hence, the required four more rational numbers are 
\(\frac { -15 }{ 25 } ,\frac { -18 }{ 30 } ,\frac { -21 }{ 35 } \)and \(\frac { -24 }{ 40 } \).

We have, \(\frac { 5 }{ 4 } =\frac { \boxed { } }{ 16 } =\frac { 25 }{ \boxed { } } =\frac { -15 }{ \boxed { } } \)
Here, the rational numbers equivalent to rational number \(\frac { 5 }{ 4 } \) are as follows: \(\frac { 5 }{ 4 } =\frac { 5\times 4 }{ 4\times 4 } =\frac { 20 }{ 16 } \)
[\(\because \) 4\(\times \)4=16, since given denominators of equivalent rational number, so multiply by 4]
\(\frac { 5 }{ 4 } =\frac { 5\times 5 }{ 4\times 5 } =\frac { 25 }{ 20 } \) 
[\(\because \) 5\(\times \)5=25, since given numerators of equivalent rational number, so multiply by 5]
Also, \(\\ \frac { 5 }{ 4 } =\frac { 5\times (-3) }{ 4\times (-3) } =\frac { -15 }{ -12 } \) 
[\(\because \) 5\(\times \)(-3)=-15, since given numerators of equivalent rational numbers, so multiply by (-3)]
So, the missing integers in the boxes are filled as 
 \(\frac { 5 }{ 4 } =\frac { \boxed { 20 } }{ 16 } =\frac { 25 }{ \boxed { 20 } } =\frac { -15 }{ \boxed { -12 } } \)

Given rational numbers are \(\frac { 1 }{ 3 } \) and  \(\frac { 1 }{ 4 } \).
LCM of 3 and 4 is 12,
\(\frac { 1\times 4 }{ 3\times 4 } =\frac { 4 }{ 12 } \) and \(\frac { 1\times 3 }{ 4\times 3 } =\frac { 3 }{ 12 } \)
Now, \(\frac { 4\times 10 }{ 12\times 10 } =\frac { 40 }{ 120 } \) and \(\frac { 3\times 10 }{ 12\times 10 } =\frac { 30 }{ 120 } \)
\(\frac { 30 }{ 120 } <\frac { 31 }{ 120 } <\frac { 32 }{ 120 } <\frac { 33 }{ 120 } <\frac { 34 }{ 120 } <\frac { 35 }{ 120 } <.......<\frac { 40 }{ 120 } \)
Hence, 5 rational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 4 } \) are
\(\frac { 31 }{ 120 } ,\frac { 32 }{ 120 } ,\frac { 33 }{ 120 } ,\frac { 34 }{ 120 } ,\frac { 35 }{ 120 } .\)

We have,\(\frac { 2 }{ 4 } +\frac { 2 }{ 10 } \)
LCM of 4 and 10 is 20 
\(\\ \frac { 2\times 5 }{ 4\times 5 } =\frac { 10 }{ 20 } ,\frac { 2\times 2 }{ 10\times 2 } =\frac { 4 }{ 20 } \)
So, \(\\ \frac { 10 }{ 20 } +\frac { 4 }{ 20 } =\frac { 10+4 }{ 20 } =\frac { 14 }{ 20 } =\frac { 7 }{ 10 } \)

We have,\(\frac { 1 }{ 3 } +\frac { 2 }{ 6 } \)
LCM of 3 and 6 is 6,
\(\frac { 1\times 2 }{ 3\times 2 } =\frac { 2 }{ 6 } \Rightarrow \frac { 2\times 1 }{ 6\times 1 } =\frac { 2 }{ 6 } \)
So,\(\frac { 2 }{ 6 } +\frac { 2 }{ 6 } =\frac { 2+2 }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Given rational numbers are \(\frac { -5 }{ 12 } \) and \(\frac { 7 }{ -18 } \) 
For the same /common denominator,
LCM of 12 and 18 is 36,
\(\frac { (-5)\times 3 }{ 12\times 3 }= \frac { -15 }{ 36 } \) and \(\frac { 7\times 2 }{ (-18\times 2) } =\frac { 14 }{ -36 }= \frac { -14 }{ 36 } \)
-14 is greater than -15.
So,\(\frac { -14 }{ 36 } >\frac { -15 }{ 36 } \)  or \(\frac { -7 }{ 18 } >\frac { -15 }{ 12 } \)

Representation of rational number\(\frac { -6 }{ 7 } \) on number line .

For common denominators, LCM of 5 and 3 is 15.
\(\frac { -4\times 3 }{ 5\times 3 } =\frac { -12 }{ 15 } \) and \(\frac { -2\times 5 }{ 3\times 5 } \frac { -10 }{ 15 } \) 
\(\frac { -12\times 2 }{ 15\times 2 } =\frac { -24 }{ 30 } \) and \(\frac { -10\times 2 }{ 15\times 2 } =\frac { -20 }{ 30 } \) 
so,\(\\ \frac { -24 }{ 30 } <\frac { -23 }{ 30 } <\frac { -22 }{ 30 } <\frac { -21 }{ 30 } <\frac { -20 }{ 30 } \) 
Hence rational numbers between \(\frac { -4 }{ 5 } \) and \(\frac { -2 }{ 3 } \)
are \(\frac { -23 }{ 30 } <\frac { -22 }{ 30 } <\frac { -21 }{ 30 }\)

Given rational numbers are \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 3 } \).
For common denominator, LCM of 2and 3 is 6,
\(\frac { 1\times 3 }{ 2\times 3 } \frac { 3 }{ 6 } \) and \(\frac { 1\times 2 }{ 3\times 2 } =\frac { 2 }{ 6 } \)
Now,\(\frac { 3\times 10 }{ 6\times 10 } =\frac { 30 }{ 60 } \) and \(\frac { 2\times 10 }{ 6\times 10 } =\frac { 20 }{ 60 } \)
so, \(\frac { 20 }{ 60 } <\frac { 21 }{ 60 } <\frac { 22 }{ 60 } <\frac { 23 }{ 60 } <.......<\frac { 30 }{ 60 } \)
Hence, 3 rational numbers between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 3 } \) are
 \(\frac { 21 }{ 60 } <\frac { 22 }{ 60 } <\frac { 23 }{ 60 }\)

\(\frac { 3 }{ 5 } \)

\(\frac { 1 }{ 3 } \)

\(\frac { 137 }{ 120 } \) 

5\(\frac { 1 }{ 12 } \)