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Let's solve these questions one by one.

### (i) \(\frac{1}{4} \text{ of } 2 \frac{2}{7} - \frac{3}{5}\)

- Convert \(2 \frac{2}{7}\) to an improper fraction:  
  \(2 \frac{2}{7} = \frac{16}{7}\)

- So the expression becomes:  
  \(\frac{1}{4} \times \frac{16}{7} - \frac{3}{5}\)

- Multiply:  
  \(\frac{1}{4} \times \frac{16}{7} = \frac{16}{28} = \frac{4}{7}\)

- Now subtract:  
  \(\frac{4}{7} - \frac{3}{5}\)

- Find the common denominator (35):  
  \(\frac{4}{7} = \frac{20}{35}, \quad \frac{3}{5} = \frac{21}{35}\)

- Subtract:  
  \(\frac{20}{35} - \frac{21}{35} = -\frac{1}{35}\)

### Answer for (i): \(-\frac{1}{35}\)

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### (ii) \(\frac{1}{4} \times \frac{1}{2} + \frac{1}{3}\)

- Multiply:  
  \(\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\)

- Now add:  
  \(\frac{1}{8} + \frac{1}{3}\)

- Find the common denominator (24):  
  \(\frac{1}{8} = \frac{3}{24}, \quad \frac{1}{3} = \frac{8}{24}\)

- Add:  
  \(\frac{3}{24} + \frac{8}{24} = \frac{11}{24}\)

### Answer for (ii): \(\frac{11}{24}\)

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### (iii) \(6 \times \frac{1}{7} \times 0 \times \frac{5}{8}\)

- The presence of \(0\) makes the entire product \(0\).

### Answer for (iii): \(0\)

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### (iv) \(\frac{3}{4} \times \frac{1}{3} \div \frac{3}{7} \times \frac{2}{5} \div \frac{5}{8}\)

- First, resolve the divisions:  
  \(\frac{1}{3} \div \frac{3}{7} = \frac{1}{3} \times \frac{7}{3} = \frac{7}{9}\)

  \(\frac{2}{5} \div \frac{5}{8} = \frac{2}{5} \times \frac{8}{5} = \frac{16}{25}\)

- Now multiply:  
  \(\frac{3}{4} \times \frac{7}{9} \times \frac{16}{25} = \frac{3 \times 7 \times 16}{4 \times 9 \times 25} = \frac{336}{900} = \frac{28}{75}\)

### Answer for (iv): \(\frac{28}{75}\)

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### (v) \(\frac{2}{4} + \frac{2}{7} \div 1 \frac{1}{3} \times \frac{2}{3}\)

- First, simplify the mixed fraction \(1 \frac{1}{3} = \frac{4}{3}\)

- Now resolve the division:  
  \(\frac{2}{7} \div \frac{4}{3} = \frac{2}{7} \times \frac{3}{4} = \frac{6}{28} = \frac{3}{14}\)

- Now the expression becomes:  
  \(\frac{2}{4} + \frac{3}{14} \times \frac{2}{3}\)

- Multiply:  
  \(\frac{3}{14} \times \frac{2}{3} = \frac{6}{42} = \frac{1}{7}\)

- Now add:  
  \(\frac{2}{4} + \frac{1}{7} = \frac{1}{2} + \frac{1}{7}\)

- Find the common denominator (14):  
  \(\frac{1}{2} = \frac{7}{14}\)

- Add:  
  \(\frac{7}{14} + \frac{1}{14} = \frac{8}{14} = \frac{4}{7}\)

### Answer for (v): \(\frac{4}{7}\)

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### (vi) \(\left(\frac{3}{7} - \frac{1}{2}\right) \text{ of } \frac{1}{7}\)

- Subtract:  
  \(\frac{3}{7} - \frac{1}{2}\)

- Find the common denominator (14):  
  \(\frac{3}{7} = \frac{6}{14}, \quad \frac{1}{2} = \frac{7}{14}\)

- Subtract:  
  \(\frac{6}{14} - \frac{7}{14} = -\frac{1}{14}\)

- Now multiply by \(\frac{1}{7}\):  
  \(-\frac{1}{14} \times \frac{1}{7} = -\frac{1}{98}\)

### Answer for

Please solve the following math problems: 
(i) 1/4 of 2 2/7 - 3/5 
(ii) 1/4 × 1/2 + 1/3 
(iii) 6 × 1/7 × 0 × 5/8 
(iv) 3/4 × 1/3 ÷ 3/7 × 2/5 ÷ 5/8 
(v) 2/4 + 2/7 ÷ 1 1/3 × 2/3 
(vi) (3/7 - 1/2) of 1/7 
(vii) (7/8 ÷ 1 1/2) of (8 1/3 ÷ 1 1/2) 
(viii) 1/3 of 60 ÷ 60

Step-by-step solutions to fraction problems involving multiplication, division, and subtraction. These problems require understanding of fraction operations and order of operations. Suitable for students in Grades 5-8.

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