Let's solve the given equations step by step.

### Equation 1:
\[ \frac{11}{15} - \left(\frac{1}{7} + x\right) \cdot \frac{3}{8} = \frac{61}{90} + \frac{x}{3} \]

First, simplify the left-hand side:
\[ \left(\frac{1}{7} + x\right) \cdot \frac{3}{8} = \frac{3}{56} + \frac{3x}{8} \]

So the equation becomes:
\[ \frac{11}{15} - \left(\frac{3}{56} + \frac{3x}{8}\right) = \frac{61}{90} + \frac{x}{3} \]

Combine the terms on the left-hand side:
\[ \frac{11}{15} - \frac{3}{56} - \frac{3x}{8} = \frac{61}{90} + \frac{x}{3} \]

Find a common denominator for the fractions on the left-hand side. The least common multiple of 15 and 56 is 840:
\[ \frac{11}{15} = \frac{11 \cdot 56}{15 \cdot 56} = \frac{616}{840} \]
\[ \frac{3}{56} = \frac{3 \cdot 15}{56 \cdot 15} = \frac{45}{840} \]

So the equation becomes:
\[ \frac{616}{840} - \frac{45}{840} - \frac{3x}{8} = \frac{61}{90} + \frac{x}{3} \]

Simplify the left-hand side:
\[ \frac{616 - 45}{840} - \frac{3x}{8} = \frac{61}{90} + \frac{x}{3} \]
\[ \frac{571}{840} - \frac{3x}{8} = \frac{61}{90} + \frac{x}{3} \]

Next, find a common denominator for the fractions on the right-hand side. The least common multiple of 90 and 3 is 90:
\[ \frac{x}{3} = \frac{x \cdot 30}{3 \cdot 30} = \frac{30x}{90} \]

So the equation becomes:
\[ \frac{571}{840} - \frac{3x}{8} = \frac{61 + 30x}{90} \]

To eliminate the fractions, find a common denominator for the entire equation. The least common multiple of 840 and 90 is 2520:
\[ \frac{571}{840} = \frac{571 \cdot 3}{840 \cdot 3} = \frac{1713}{2520} \]
\[ \frac{3x}{8} = \frac{3x \cdot 315}{8 \cdot 315} = \frac{945x}{2520} \]
\[ \frac{61 + 30x}{90} = \frac{61 \cdot 28 + 30x \cdot 28}{90 \cdot 28} = \frac{1708 + 840x}{2520} \]

So the equation becomes:
\[ \frac{1713}{2520} - \frac{945x}{2520} = \frac{1708 + 840x}{2520} \]

Combine the terms:
\[ 1713 - 945x = 1708 + 840x \]

Solve for \( x \):
\[ 1713 - 1708 = 840x + 945x \]
\[ 5 = 1785x \]
\[ x = \frac{5}{1785} \]
\[ x = \frac{1}{357} \]

### Equation 2:
\[ 4x - (2x + 1) = \frac{10}{3} + x \]

Simplify the left-hand side:
\[ 4x - 2x - 1 = \frac{10}{3} + x \]
\[ 2x - 1 = \frac{10}{3} + x \]

Subtract \( x \) from both sides:
\[ 2x - x - 1 = \frac{10}{3} \]
\[ x - 1 = \frac{10}{3} \]

Add 1 to both sides:
\[ x = \frac{10}{3} + 1 \]
\[ x = \frac{10}{3} + \frac{3}{3} \]
\[ x = \frac{13}{3} \]

So the solutions are:
\[ x = \frac{1}{357} \text{ (from Equation 1)} \]
\[ x = \frac{13}{3} \text{ (from Equation 2)} \]