1) \( \left(\sqrt[3]{\frac{1}{3}}\right)^{x+6}>\frac{1}{9} \)

To solve the inequality \[ \left(\sqrt[3]{\frac{1}{3}}\right)^{x+6} > \frac{1}{9}, \] we start by rewriting the left-hand side. The expression \(\sqrt[3]{\frac{1}{3}}\) can be expressed as: \[ \sqrt[3]{\frac{1}{3}} = \left(\frac{1}{3}\right)^{1/3} = 3^{-1/3}. \] Thus, we can rewrite the left-hand side of the inequality: \[ \left(\sqrt[3]{\frac{1}{3}}\right)^{x+6} = \left(3^{-1/3}\right)^{x+6} = 3^{-\frac{1}{3}(x+6)}. \] Now, we rewrite the right-hand side \(\frac{1}{9}\): \[ \frac{1}{9} = 3^{-2}. \] Now, we can rewrite the inequality as: \[ 3^{-\frac{1}{3}(x+6)} > 3^{-2}. \] Since the base \(3\) is positive and greater than \(1\), we can compare the exponents directly: \[ -\frac{1}{3}(x+6) > -2. \] To eliminate the negative sign, we multiply both sides by \(-1\) and reverse the inequality: \[ \frac{1}{3}(x+6) < 2. \] Next, we multiply both sides by \(3\): \[ x + 6 < 6. \] Now, we subtract \(6\) from both sides: \[ x < 0. \] Thus, the solution to the inequality is: \[ \boxed{(-\infty, 0)}. \]