To solve the given problems, let's break them down step by step.

### a) Calculate \( A \)

\[ A = 50 - \left( \frac{4}{13} + \frac{2}{15} + \frac{2}{17} \right) \div \left( 100 - \left( \frac{8}{13} + \frac{4}{15} + \frac{4}{17} \right) \right) \]

First, calculate the fractions inside the parentheses:

\[ \frac{4}{13} + \frac{2}{15} + \frac{2}{17} \]

To add these fractions, find a common denominator, which is \( 13 \times 15 \times 17 = 3315 \).

\[ \frac{4}{13} = \frac{4 \times 255}{3315} = \frac{1020}{3315} \]
\[ \frac{2}{15} = \frac{2 \times 221}{3315} = \frac{442}{3315} \]
\[ \frac{2}{17} = \frac{2 \times 195}{3315} = \frac{390}{3315} \]

Adding these fractions:

\[ \frac{1020 + 442 + 390}{3315} = \frac{1852}{3315} \]

Next, calculate the second set of fractions:

\[ \frac{8}{13} + \frac{4}{15} + \frac{4}{17} \]

Using the same common denominator:

\[ \frac{8}{13} = \frac{8 \times 255}{3315} = \frac{2040}{3315} \]
\[ \frac{4}{15} = \frac{4 \times 221}{3315} = \frac{884}{3315} \]
\[ \frac{4}{17} = \frac{4 \times 195}{3315} = \frac{780}{3315} \]

Adding these fractions:

\[ \frac{2040 + 884 + 780}{3315} = \frac{3704}{3315} \]

Now, substitute back into the original equation:

\[ A = 50 - \frac{\frac{1852}{3315}}{100 - \frac{3704}{3315}} \]

Simplify the denominator:

\[ 100 - \frac{3704}{3315} = \frac{331500 - 3704}{3315} = \frac{327796}{3315} \]

Now, the equation becomes:

\[ A = 50 - \frac{1852}{327796} \]

Since \( \frac{1852}{327796} \) is a very small number, \( A \approx 50 \).

### b) Calculate \( B \)

\[ B = \frac{1}{8} + \frac{1}{24} + \frac{1}{48} + \frac{1}{80} + \frac{1}{120} \]

Find a common denominator, which is the least common multiple (LCM) of 8, 24, 48, 80, and 120. The LCM is 240.

Convert each fraction:

\[ \frac{1}{8} = \frac{30}{240} \]
\[ \frac{1}{24} = \frac{10}{240} \]
\[ \frac{1}{48} = \frac{5}{240} \]
\[ \frac{1}{80} = \frac{3}{240} \]
\[ \frac{1}{120} = \frac{2}{240} \]

Add these fractions:

\[ B = \frac{30 + 10 + 5 + 3 + 2}{240} = \frac{50}{240} = \frac{5}{24} \]

### c) Calculate \( C \)

\[ C = \frac{1}{19} + \frac{9}{19 \cdot 29} + \frac{9}{29 \cdot 39} + \ldots + \frac{9}{1999 \cdot 2009} \]

This is a telescoping series. The general term is:

\[ \frac{9}{(19 + 10k)(19 + 10(k+1))} \]

The series simplifies to:

\[ \frac{1}{19} + \left( \frac{1}{19} - \frac{1}{29} \right) + \left( \frac{1}{29} - \frac{1}{39} \right) + \ldots + \left( \frac{1}{1999} - \frac{1}{2009} \right) \]

Most terms cancel out, leaving:

\[ C = \frac{1}{19} + 1 - \frac{1}{2009} \approx 1 \]

So, the final answers are:

\[ A \approx 50 \]
\[ B = \frac{5}{24} \]
\[ C \approx 1 \]