(i)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 3 and 4 = 12

\(\frac{3}{4}= \frac{3\times3}{4\times3}=\frac{9}{12}\)

And,

\(\frac{1}{3}= \frac{1\times4}{3\times4}=\frac{4}{12}\)

Therefore,

= \(\frac{1}{3}-\frac{3}{4}\)

= \(\frac{4}{12}-\frac{9}{12}\)

= \(\frac{4-9}{12}\)

= \(\frac{-5}{12}\)

(ii)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 6 and 3 = 6

\(\frac{-5}{6}= \frac{-5\times1}{6\times1}=\frac{-5}{6}\)

And,

\(\frac{1}{3}= \frac{1\times2}{3\times2}=\frac{2}{6}\)

Therefore,

\(\frac{1}{3}-(\frac{-5}{6})\)

= \(\frac{2}{6}-(\frac{-5}{6})\)

= \(\frac{2-(-5)}{6}\)

= \(\frac{2+5}{6}\)

= \(\frac{7}{6}\)

(iii)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 9 and 5 = 45

\(\frac{-8}{9}= \frac{-8\times5}{9\times5}=\frac{-40}{45}\)

And,

\(\frac{-3}{5}= \frac{-3\times9}{5\times9}=\frac{-27}{45}\)

Therefore,

\(\frac{-3}{5}-(\frac{-8}{9})\)

= \(\frac{-27}{45}-(\frac{-40}{45})\) 

= \(\frac{-27-(-40)}{45}\) 

= \(\frac{-27+40}{45}\)

= \(\frac{13}{45}\)

(iv)

We can write, -1 = \(\frac{-1}{1}\)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 1 and 7 =7

\(\frac{-1}{1}= \frac{-1\times7}{1\times7}=\frac{-7}{7}\)

And,

\(\frac{-9}{7}= \frac{-9\times1}{7\times1}=\frac{-9}{7}\)

Therefore,

\(-1-(\frac{-9}{7})\)

= \(\frac{-7}{7}-(\frac{-9}{7})\)

= \(\frac{-7+9}{7}\)

= \(\frac{2}{7}\)