Let's solve each of these expressions step-by-step:

**a)** \( \frac{3}{4} + \frac{2}{7} + \frac{1}{4} + \frac{3}{5} + \frac{5}{7} \)

First, find a common denominator for all the fractions. The denominators are 4, 7, and 5. The least common multiple (LCM) of these numbers is 140.

Now, convert each fraction to have the denominator of 140:

- \( \frac{3}{4} = \frac{3 \times 35}{4 \times 35} = \frac{105}{140} \)
- \( \frac{2}{7} = \frac{2 \times 20}{7 \times 20} = \frac{40}{140} \)
- \( \frac{1}{4} = \frac{1 \times 35}{4 \times 35} = \frac{35}{140} \)
- \( \frac{3}{5} = \frac{3 \times 28}{5 \times 28} = \frac{84}{140} \)
- \( \frac{5}{7} = \frac{5 \times 20}{7 \times 20} = \frac{100}{140} \)

Now sum the fractions:

\[
\frac{105}{140} + \frac{40}{140} + \frac{35}{140} + \frac{84}{140} + \frac{100}{140} = \frac{364}{140}
\]

Simplify the fraction:

\[
\frac{364}{140} = \frac{182}{70} = \frac{91}{35} = \frac{13}{5}
\]

So the result of expression (a) is \(\frac{13}{5}\).

**b)** \( \frac{2}{17} + \frac{15}{23} - \frac{15}{17} - \frac{4}{19} + \frac{8}{23} \)

First, find a common denominator for these fractions. The denominators are 17, 23, and 19. The LCM is the product \(17 \times 23 \times 19 = 7429\).

Now convert each fraction to have the denominator of 7429:

- \( \frac{2}{17} = \frac{2 \times 437}{17 \times 437} = \frac{874}{7429} \)
- \( \frac{15}{23} = \frac{15 \times 323}{23 \times 323} = \frac{4845}{7429} \)
- \( \frac{15}{17} = \frac{15 \times 437}{17 \times 437} = \frac{6555}{7429} \)
- \( \frac{4}{19} = \frac{4 \times 391}{19 \times 391} = \frac{1564}{7429} \)
- \( \frac{8}{23} = \frac{8 \times 323}{23 \times 323} = \frac{2584}{7429} \)

Now sum the fractions:

\[
\left( \frac{874}{7429} + \frac{4845}{7429} + \frac{2584}{7429} \right) - \left( \frac{6555}{7429} + \frac{1564}{7429} \right)
\]

Calculate separately:

\[
\text{Positive terms: } \frac{874 + 4845 + 2584}{7429} = \frac{8303}{7429}
\]
\[
\text{Negative terms: } \frac{6555 + 1564}{7429} = \frac{8119}{7429}
\]

Now subtract the negative sum from the positive sum:

\[
\frac{8303 - 8119}{7429} = \frac{184}{7429}
\]

So the result of expression (b) is \(\frac{184}{7429}\).

**c)** \( \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{5} + \frac{1}{5} \times \frac{1}{6} \)

Calculate each product:

- \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)
- \( \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \)
- \( \frac{1}{4} \times \frac{1}{5} = \frac{1}{20} \)
- \( \frac{1}{5} \times \frac{1}{6} = \frac{1}{30} \)

Find a common denominator, which is 60:

- \( \frac{1}{4} = \frac{15}{60} \)
- \( \frac{1}{6} = \frac{10}{60} \)
- \( \frac{1}{12} = \frac{5}{60} \)
- \( \frac{1}{20} = \frac{3}{60} \)
- \( \frac{1}{30} = \frac{2}{60} \)

Sum them up:

\[
\frac{15}{60} + \frac{10}{60} + \frac{5}{60} + \frac{3}{60} + \frac{2}{60} = \frac{35}{60} = \frac{7}{12}
\]

So the result of expression (c) is \(\frac{7}{12}\).

Final results:
- a) \(\frac{13}{5}\)
- b) \(\frac{184}{7429}\)
- c) \(\frac{7}{12}\)