Let's solve the expressions for \( A \) and \( B \) step by step.

### Expression for \( A \):

\[ A = \frac{1}{3} - \left[ -\frac{5}{4} - \left( \frac{1}{4} + \frac{3}{8} \right) \right] \]

First, simplify the inner parentheses:

\[ \frac{1}{4} + \frac{3}{8} \]

To add these fractions, we need a common denominator. The least common multiple of 4 and 8 is 8.

\[ \frac{1}{4} = \frac{2}{8} \]

So,

\[ \frac{2}{8} + \frac{3}{8} = \frac{5}{8} \]

Now substitute back into the expression:

\[ A = \frac{1}{3} - \left[ -\frac{5}{4} - \frac{5}{8} \right] \]

Next, find a common denominator for \(\frac{5}{4}\) and \(\frac{5}{8}\). The least common multiple of 4 and 8 is 8.

\[ \frac{5}{4} = \frac{10}{8} \]

So,

\[ -\frac{10}{8} - \frac{5}{8} = -\frac{15}{8} \]

Now substitute back into the expression:

\[ A = \frac{1}{3} - \left( -\frac{15}{8} \right) \]

\[ A = \frac{1}{3} + \frac{15}{8} \]

To add these fractions, we need a common denominator. The least common multiple of 3 and 8 is 24.

\[ \frac{1}{3} = \frac{8}{24} \]
\[ \frac{15}{8} = \frac{45}{24} \]

So,

\[ \frac{8}{24} + \frac{45}{24} = \frac{53}{24} \]

Thus,

\[ A = \frac{53}{24} \]

### Expression for \( B \):

\[ B = \left( \frac{1}{2} - \frac{7}{13} - \frac{1}{3} \right) + \left( -\frac{6}{13} + \frac{1}{2} + 1 \frac{1}{2} \right) \]

First, simplify the mixed number \( 1 \frac{1}{2} \):

\[ 1 \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2} \]

Now, simplify the second part of the expression:

\[ -\frac{6}{13} + \frac{1}{2} + \frac{3}{2} \]

Combine \(\frac{1}{2}\) and \(\frac{3}{2}\):

\[ \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2 \]

So,

\[ -\frac{6}{13} + 2 \]

Convert 2 to a fraction with a denominator of 13:

\[ 2 = \frac{26}{13} \]

So,

\[ -\frac{6}{13} + \frac{26}{13} = \frac{20}{13} \]

Now, simplify the first part of the expression:

\[ \frac{1}{2} - \frac{7}{13} - \frac{1}{3} \]

Find a common denominator for \(\frac{1}{2}\), \(\frac{7}{13}\), and \(\frac{1}{3}\). The least common multiple of 2, 13, and 3 is 78.

\[ \frac{1}{2} = \frac{39}{78} \]
\[ \frac{7}{13} = \frac{42}{78} \]
\[ \frac{1}{3} = \frac{26}{78} \]

So,

\[ \frac{39}{78} - \frac{42}{78} - \frac{26}{78} = \frac{39 - 42 - 26}{78} = \frac{-29}{78} \]

Now combine the two parts:

\[ B = \frac{-29}{78} + \frac{20}{13} \]

Convert \(\frac{20}{13}\) to a fraction with a denominator of 78:

\[ \frac{20}{13} = \frac{120}{78} \]

So,

\[ B = \frac{-29}{78} + \frac{120}{78} = \frac{91}{78} \]

Thus,

\[ B = \frac{91}{78} \]

So the final results are:

\[ A = \frac{53}{24} \]
\[ B = \frac{91}{78} \]