To solve the inequality \((-2x + 5)(3x + \frac{2}{3}) > 0\), we will analyze the two factors:

1. \(-2x + 5\)
2. \(3x + \frac{2}{3}\)

### Step 1: Find the zeros of each factor

**Factor 1: \(-2x + 5 = 0\)**

\[
-2x + 5 = 0 \implies -2x = -5 \implies x = \frac{5}{2}
\]

**Factor 2: \(3x + \frac{2}{3} = 0\)**

\[
3x + \frac{2}{3} = 0 \implies 3x = -\frac{2}{3} \implies x = -\frac{2}{9}
\]

### Step 2: Identify critical points
The critical points from the factors are:
- \(x = \frac{5}{2}\)
- \(x = -\frac{2}{9}\)

### Step 3: Determine intervals
These critical points divide the number line into three intervals:

1. \( (-\infty, -\frac{2}{9}) \)
2. \( (-\frac{2}{9}, \frac{5}{2}) \)
3. \( (\frac{5}{2}, \infty) \)

### Step 4: Test the sign of the product in each interval

1. **Interval** \( (-\infty, -\frac{2}{9}) \):
- Choose \(x = -1\):
\[
(-2(-1) + 5)(3(-1) + \frac{2}{3}) = (2 + 5)(-3 + \frac{2}{3}) = 7(-\frac{7}{3}) < 0
\]

2. **Interval** \( (-\frac{2}{9}, \frac{5}{2}) \):
- Choose \(x = 0\):
\[
(-2(0) + 5)(3(0) + \frac{2}{3}) = (5)(\frac{2}{3}) = \frac{10}{3} > 0
\]

3. **Interval** \( (\frac{5}{2}, \infty) \):
- Choose \(x = 3\):
\[
(-2(3) + 5)(3(3) + \frac{2}{3}) = (-6 + 5)(9 + \frac{2}{3}) = (-1)(\frac{29}{3}) < 0
\]

### Step 5: Conclusion
The product \((-2x + 5)(3x + \frac{2}{3}) > 0\) is satisfied in the interval \( (-\frac{2}{9}, \frac{5}{2}) \).

Thus, the solution to the inequality is:

\[
\boxed{(-\frac{2}{9}, \frac{5}{2})}
\]

Note that we do not include the endpoints \( -\frac{2}{9} \) and \( \frac{5}{2} \) because the inequality is strict (> 0).