To solve these equations, we'll use the property of exponents that states if \( a^m \times a^n = a^p \), then \( m + n = p \). We'll solve for \( x \) in each case:

1) \(\left( \frac{3}{4} \right)^{5} \cdot x = \left( \frac{3}{4} \right)^{7}\)

- \(5 + n = 7\)
- \(n = 7 - 5 = 2\)
- \(x = \left( \frac{3}{4} \right)^{2}\)

2) \(\left( \frac{2}{3} \right)^{2} \cdot x = \left( \frac{2}{3} \right)^{4}\)

- \(2 + n = 4\)
- \(n = 4 - 2 = 2\)
- \(x = \left( \frac{2}{3} \right)^{2}\)

3) \(\left( -\frac{2}{5} \right)^{2} \cdot x = \left( -\frac{2}{5} \right)^{4}\)

- \(2 + n = 4\)
- \(n = 4 - 2 = 2\)
- \(x = \left( -\frac{2}{5} \right)^{2}\)

4) \(x \cdot \left( \frac{2}{3} \right) = \left( \frac{2}{3} \right)^{2}\)

- \(n + 1 = 2\)
- \(n = 2 - 1 = 1\)
- \(x = \left( \frac{2}{3} \right)^{1} = \frac{2}{3}\)

5) \(x \cdot \left( -\frac{1}{2} \right)^{3} = -\frac{1}{2}\)

- \(n + 3 = 1\)
- \(n = 1 - 3 = -2\)
- \(x = \left( -\frac{1}{2} \right)^{-2} = \left( -\frac{1}{2} \right)^{-2} = \left( -\frac{1}{2} \right)^{-2}\)

6) \(\left( -\frac{9}{5} \right)^{7} \cdot x = \left( -\frac{9}{5} \right)^{8}\)

- \(7 + n = 8\)
- \(n = 8 - 7 = 1\)
- \(x = \left( -\frac{9}{5} \right)^{1} = -\frac{9}{5}\)

7) \(\left( -\frac{5}{9} \right)^{10} \cdot x = \left( -\frac{5}{9} \right)^{8}\)

- \(10 + n = 8\)
- \(n = 8 - 10 = -2\)
- \(x = \left( -\frac{5}{9} \right)^{-2}\)

8) \(\left( -\frac{1}{3} \right)^{2} \cdot x = \left( -\frac{1}{3} \right)^{4}\)

- \(2 + n = 4\)
- \(n = 4 - 2 = 2\)
- \(x = \left( -\frac{1}{3} \right)^{2}\)

9) \(\left( \frac{4}{5} \right)^{5} \cdot x = \left( \frac{-4}{5} \right)^{4}\)

- \(5 + n = 4\)
- \(n = 4 - 5 = -1\)
- Note: The given equality \(\left( \frac{4}{5} \right)^{5} \cdot x = \left( -\frac{4}{5} \right)^{4}\) suggests there may be more to consider here, as the bases have different signs. Normally, we'd need context or additional clarification for such cases, but we'll proceed given what's provided.
- \(x = \left( \frac{4}{5} \right)^{-1} = \frac{5}{4}\)

Each solution for \(x\) maintains the equality of the expressions.