1. **Find the additive inverse of $-\frac{2}{3}$**. The additive inverse of a number $a$ is the number that when added to $a$ gives zero. Formula: $a + (-a) = 0$ Given $a = -\frac{2}{3}$, the additive inverse is $-a = -\left(-\frac{2}{3}\right) = \frac{2}{3}$. Check: $-\frac{2}{3} + \frac{2}{3} = 0$ 2. **Find the multiplicative inverse of $-\frac{2}{3}$**. The multiplicative inverse of a number $a$ (except zero) is the number that when multiplied by $a$ gives one. Formula: $a \times a^{-1} = 1$ Given $a = -\frac{2}{3}$, the multiplicative inverse is $a^{-1} = -\frac{3}{2}$. Check: $-\frac{2}{3} \times -\frac{3}{2} = \frac{6}{6} = 1$ 3. **Determine for which values of $x \in \{-6, -4, -3, 2, 3\}$ the inequality $x > -3$ is true**. Check each value: - $-6 > -3$? No - $-4 > -3$? No - $-3 > -3$? No (equal, not greater) - $2 > -3$? Yes - $3 > -3$? Yes So the values are $\{2, 3\}$. 4. **Write all integers less than 5 in set builder notation**. The integers less than 5 are $\{..., -3, -2, -1, 0, 1, 2, 3, 4\}$. Set builder notation: $\{x \mid x \in \mathbb{Z}, x < 5\}$. **Final answers:** - Additive inverse of $-\frac{2}{3}$ is $\boxed{\frac{2}{3}}$. - Multiplicative inverse of $-\frac{2}{3}$ is $\boxed{-\frac{3}{2}}$. - Values of $x$ such that $x > -3$ are $\boxed{\{2, 3\}}$. - Integers less than 5 in set builder notation: $\boxed{\{x \mid x \in \mathbb{Z}, x < 5\}}$.