The complete boxes are shown below:
\(\frac{1}{7}\times \frac{1}{5}=\boxed {\frac{1\times 1}{7\times 5}} = \boxed {\frac{1}{35}}\)

Mixed fraction =Quotient \(\frac{Reminder}{Divisor}=4 \frac{1}{3}\)

Solved part of exercise by Rohan = \(\frac{3}{7}\)       [given]
Solved parr of exercise by Ramu = \(\frac{2}{5}\)    [given]
Let us compare \(\frac{3}{7}\)and \(\frac{2}{5}\)
LCM is 7 and 5 is 35.
\(\frac{3}{7}=\frac{3}{7}\times \frac{5}{5}=\frac{15}{35}\) and \(\frac{2}{5}=\frac{2}{5}\times \frac{7}{7}=\frac{14}{35}\). Thus, \(\frac{15}{35}>\frac{14}{35}\) or \(\frac{3}{7}>\frac{2}{5}\)
So, Ramu solved lesser part i.e. \(\frac{2}{5}\).

Rs 1208.7

(a) We have, \(\frac{1}{4}\)of \(\frac{1}{4}\)=\(\frac{1}{4}\times \frac{1}{4}=\frac{1\times 1}{4\times 4}=\frac{1}{16}\)  [∵'of' means multiplication]
(b) We have \(\frac{1}{4}\) of \(\frac{3}{5} \)=\(\frac{1}{4}\times \frac{3}{5}= \frac{1\times 3}{4 \times 5}= \frac{3}{20}\)
(c) We have  \(\frac{1}{4}\) of \(\frac{4}{3}\)=\(\frac{1}{4}\times \frac{4}{3}=\frac{1\times 4}{4 \times 3}=\frac{4}{12}\) [dividing numerator and denominator by 4]
=\(\frac{1}{3}\)

Let an apple be 1 part. Rim ate \(\frac{3}{5}\)part of an apple.
Now, remaining part=1-\(\frac{3}{5}\)=\(\frac{1}{1}-\frac{3}{5}=\frac{1\times 5-3\times 1}{5}\)
=\(\frac{5-3}{5}=\frac{2}{5}\)
So, Somu ate 2/5 part of an apple.
∵ 3>2  [numerators of fractions]
∴ \(\frac{3}{5}\)>\(\frac{2}{5}\)
i.e. Riru had larger share.
Now, d'ifference between theeilr shares=\(\frac{3}{5}\)-\(\frac{2}{5}\)=\(\frac{3-2}{5}=\frac{1}{5}\)
Hence, Ritu had the larger share which is \(\frac{1}{5}\) more than Sornu's share.

(i) We have, \(3\times \frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow \frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\)
It shows that, on combining three figures, in each of them out of 5 parts, 1 part is shaded. It is represented by the figure (c),
(ii) We have, \(2\times \frac{1}{3}=\frac{2}{3}\Rightarrow ​​\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
It shows that on combining two figures, in each of them out of 3 parts, 1 part is shaded, It is represented by the figure (a),
(iii) We have, \(3\times \frac{3}{4}=2\frac{1}{4}\)
\(\Rightarrow 3\times \frac{3}{4}=\frac{9}{4} \Rightarrow \frac{3}{4}+\frac{3}{4}+\frac{3}{4}=\frac{9}{4}\)
It shows that, on combining three figures, in each of them out of 4 parts, 3 parts are shaded, We get two figures in which all 4 parts are shaded and in third figure out of 4 parts, 1 part is shaded and it is represented by the figure (b)