We have
\(\frac{3}{7}=\frac{3\times5}{7\times5}=\frac{15}{35}\)
\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35} \)
\(\frac{28}{35}>\frac{15}{35}\Rightarrow\frac{4}{5}>\frac{3}{7}\)
\(\Rightarrow \frac{-4}{5}<\frac{-3}{7}\Rightarrow \frac{-3}{7}>(-\frac{4}{5})\)

We have
\(\frac{2}{9}=\frac{2\times5}{9\times5}=\frac{10}{45}\)
\(\frac{3}{5}=\frac{3\times9}{5\times9}=\frac{27}{45}\)
\(\frac{27}{45}>\frac{10}{45}\Rightarrow \frac{3}{5}>\frac{2}9{}\)
\(\Rightarrow \frac{2}{9}<\frac{3}{5}\Rightarrow-\frac{2}{9}>(-\frac{3}{5})\)

(a) We have,
\(\frac{2}{3}\times\frac{-7}{8}=\frac{1\times(-7)}{3\times4}=\frac{-7}{12}\)
(b) We have,
\(\frac{-6}{7}\times\frac{5}{7}\)=\(\frac{(-6)\times5}{7\times7}=\frac{-30}{49}\)

Points D on the number line represents the rational number \(\)7/8 as shown below.

We have,\(-\left( 2\frac { 1 }{ 3 } \right) +\left( 4\frac { 3 }{ 5 } \right) =-\left( \frac { 2\times 3+1 }{ 3 } \right) +\frac { 4\times 3+5 }{ 5 } =\frac { -7 }{ 3 } +\frac { 23 }{ 5 } \)
\(\because \) LCM of 3 and 5 =15
\(\therefore \) \(\frac { -7 }{ 3 } =\frac { -7\times 5 }{ 3\times 5 } =\frac { -35 }{ 15 } \) and \(\frac { 23 }{ 5 } =\frac { 23\times 3 }{ 5\times 3 } =\frac { 69 }{ 15 } \)
Now, \(-2\frac { 1 }{ 3 } +4\frac { 3 }{ 5 } =\frac { -35 }{ 15 } +\frac { 69 }{ 15 } =\frac { -35+69 }{ 15 } =\frac { -34 }{ 15 } =2\frac { 4 }{ 15 } \)

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,\(\frac { -1 }{ 3 } ,\frac { -2 }{ 9 } ,\frac { -4 }{ 3 } \)
Here, the denominators of the given rational numbers are positive.      

\(\because \) LCM = 3\(\times \)3=9
\(\therefore \) \(\frac { -1 }{ 3 } =\frac { -1\times 3 }{ 3\times 3 } =\frac { -3 }{ 9 } ,\)
\(\frac { -2 }{ 9 } =\frac { -2\times 1 }{ 9\times 1 } =\frac { -2 }{ 9 } \)
 and \(\frac { -4 }{ 3 } =\frac { -4\times 3 }{ 3\times 3 } =\frac { -12 }{ 9 } \)
We observe that, 3 < 2<12
\(\Rightarrow \frac { 2 }{ 9 } <\frac { 3 }{ 9 } <\frac { 12 }{ 9 } \Rightarrow \frac { -2 }{ 9 } >\frac { -3 }{ 9 } >\frac { -12 }{ 9 }\)
\( \\ \Rightarrow \frac { -2 }{ 9 } >\frac { -1 }{ 3 } >\frac { -4 }{ 3 } \)
Hence, the given rational number when arranged in ascending order are \(\frac { -4 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { -2 }{ 9 }.\)

\(\\ \frac { -8 }{ 5 } >\frac { -7 }{ 4 } \)

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 \(\frac { -24 }{ 40 } \).

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

Given, \(\frac{-3}{5} \) as a rational number
with numerator  -15,
\(\frac { -3\times 5 }{ 5\times 5 } =\frac { -15 }{ 25 } \)

Given ,  \(\frac { 5}{ -6 } \)
Rational number with positive denominator is \(\frac { 5}{ 6 } \)

Given, \(\frac { -4 }{ -5 } \)
Rational number with positive denominator is \(\frac { 4 }{ 5 } \)

Given numerator and denominator are -2 and - 7.
 So, rational number will be \(\frac { -2 }{ -7 } \) i.e.\(\frac { 2 }{ 7 } \)

Given numerator and denominator are 3 and 5.
 So, rational number will be \(\frac { 3 }{ 5 } \)

\(\frac { 75 }{ 182 } \)

\(\frac { -3 }{ 65 } \)

\(\frac { 95 }{ 168 } \)

\(\frac { 24 }{ 5 } \)

-\(\frac { 3 }{ 4 } \)