Let's solve each part step by step.

### Part a:
\[ \frac{7}{13} \times \frac{7}{15} - \frac{5}{12} \times \frac{21}{39} + \frac{49}{91} \times \frac{8}{15} \]

1. Calculate \(\frac{7}{13} \times \frac{7}{15}\):
\[ \frac{7 \times 7}{13 \times 15} = \frac{49}{195} \]

2. Calculate \(\frac{5}{12} \times \frac{21}{39}\):
\[ \frac{5 \times 21}{12 \times 39} = \frac{105}{468} \]
Simplify \(\frac{105}{468}\):
\[ \frac{105 \div 3}{468 \div 3} = \frac{35}{156} \]

3. Calculate \(\frac{49}{91} \times \frac{8}{15}\):
\[ \frac{49 \times 8}{91 \times 15} = \frac{392}{1365} \]

Now, combine these results:
\[ \frac{49}{195} - \frac{35}{156} + \frac{392}{1365} \]

To combine these fractions, find a common denominator. The least common multiple (LCM) of 195, 156, and 1365 is 40920.

Convert each fraction to have the common denominator 40920:
\[ \frac{49}{195} = \frac{49 \times 210}{195 \times 210} = \frac{10290}{40920} \]
\[ \frac{35}{156} = \frac{35 \times 262}{156 \times 262} = \frac{9170}{40920} \]
\[ \frac{392}{1365} = \frac{392 \times 30}{1365 \times 30} = \frac{11760}{40920} \]

Now, perform the operations:
\[ \frac{10290}{40920} - \frac{9170}{40920} + \frac{11760}{40920} \]
\[ = \frac{10290 - 9170 + 11760}{40920} \]
\[ = \frac{13880}{40920} \]

Simplify the fraction:
\[ \frac{13880 \div 20}{40920 \div 20} = \frac{694}{2046} \]
\[ \frac{694 \div 2}{2046 \div 2} = \frac{347}{1023} \]

So, the result for part a is:
\[ \frac{347}{1023} \]

### Part b:
\[ \left( \frac{12}{199} + \frac{23}{200} - \frac{34}{201} \right) \times \left( \frac{1}{2} - \frac{1}{3} - \frac{1}{6} \right) \]

1. Calculate \(\frac{1}{2} - \frac{1}{3} - \frac{1}{6}\):
\[ \frac{1}{2} - \frac{1}{3} - \frac{1}{6} = \frac{3}{6} - \frac{2}{6} - \frac{1}{6} = \frac{3 - 2 - 1}{6} = \frac{0}{6} = 0 \]

Since one of the factors is 0, the entire expression evaluates to 0.

So, the result for part b is:
\[ 0 \]

### Summary:
a. \(\frac{347}{1023}\)
b. \(0\)