Since there is no rational number which when multiplied by 0 gives 1, therefore, zero has no reciprocal.

\(\frac{-2}{7}+0=0+(\frac{-2}{7})=\frac{-2}{7}\).

\(\frac{-5}{4} \div \frac{3}{7} = \frac{-35}{12}\)
and \(\frac{3}{7} \div \frac{-5}{4} = \frac{-12}{35}\) 
So \(\frac{-5}{4} \div \frac{3}{7} \ne \frac{3}{7} \div (\frac{-5}{4})\).

Yes; 8 + 5 = 13,
which is an integer.

\(\therefore\) product of two rational numbers  = \(\frac { -28 }{ 75 } \)
Any one of the ratioanl numbers = \(\frac { 14 }{ 25 } \)
\(\therefore\) the other number = \(\left[ \frac { -28 }{ 75 } \right] \div \frac { 14 }{ 25 } \)
= \(\frac { -28 }{ 75 } \times \frac { 25 }{ 14 } =\frac { -2\times 1 }{ 3\times 1 } =\frac { -2 }{ 3 } \)
Thus, the required rational number is \(\left( \frac { -2 }{ 3 } \right) \)

\(\therefore\) Reciprocal of  \(\frac { 7 }{ 8 } \) is \(\frac { 8 }{ 7 } \)
Reciprocal  of \(\frac { -2 }{ 21 } \) is \(\frac { -21 }{ 2 } \)
\(\therefore\) \(\left[ Reciprocal\ of\frac { 7 }{ 8 } \right] \times \left[ Reciprocal\ of\left( \frac { -2 }{ 21 } \right) \right] \)
= \(\frac { 8 }{ 7 } \times \left( \frac { -21 }{ 2 } \right) =\frac { 4\times (-3) }{ 1\times 1 } =-12\)

\({-13\over15},{-12\over15},{-11\over15},{-10\over15},{-9\over15}\)

The additive Inverse of \({-11\over 13}\) is \({11\over 13}\)
The multiplicative Inverse of \({-11\over 13}\) is \({-13\over 11}\)
So, required product = \({11\over 13} \times {-13\over11}=-1\) 

The additive Inverse of \({7\over3}\)is \({-7\over3}\)
and muItiplication Inverse of  \({7\over3}\)is \({3\over7}\)
So, required sum = \({-7\over3}+{3\over7}={-49+9\over21}={-40\over21}\)

For rational numbers \({5\over4},{3\over4}and{1\over4}\)  we see that
\({5\over4}-({3\over4}-{1\over4})={5\over4}-{2\over4}={3\over4}\) and \(({5\over4}-{3\over4})-{1\over4}={2\over4}-{1\over4}={1\over4}\)
Thus, \({5\over4}-({3\over4}-{1\over4})\neq ({5\over4}-{3\over4})-{1\over4}\)
So, we can say that subtraction is not associative for rational numbers.

\({-19\over24},{-18\over24},{-17\over24},{-16\over24},{-15\over24}{}\)

There can be infinitely many rational numbers smaller than 2. Five of them are \(1,{1\over2},0,-1,-{1\over2}\)

No, here \({8\over9}\times -1{1\over8}={8\over9}\times{-9\over8}=-1\) which is not equal to 1.Therercore, \({8\over9}\)  is not multiplicative inverse of \(-1{1\over8}\)

we have, \(1\over5\)
\(\therefore\)The multiplicative inverse of \(1\over5\) is 5

we have, \(-5\over9\) so, additive inverse of \(-5\over9\) is \(5\over9\)