1. We are given the inequality $\left(\frac{2}{3}\right)^{x-2} > \frac{9}{4}$. We want to find the values of $x$ that satisfy this. 2. Recall that $\left(\frac{2}{3}\right)^{x-2} = \frac{2^{x-2}}{3^{x-2}}$ and $\frac{9}{4} = \left(\frac{3}{2}\right)^2$. Since $\frac{2}{3} < 1$, the function $\left(\frac{2}{3}\right)^t$ is decreasing in $t$. 3. Rewrite the inequality as: $$\left(\frac{2}{3}\right)^{x-2} > \left(\frac{3}{2}\right)^2$$ 4. Because $\left(\frac{2}{3}\right)^{x-2} = \left(\frac{3}{2}\right)^{-(x-2)}$, the inequality becomes: $$\left(\frac{3}{2}\right)^{-(x-2)} > \left(\frac{3}{2}\right)^2$$ 5. Since $\frac{3}{2} > 1$, the function $\left(\frac{3}{2}\right)^t$ is increasing, so the inequality $$-(x-2) > 2$$ must hold. 6. Simplify: $$-x + 2 > 2$$ $$-x > 0$$ $$\cancel{-}x > \cancel{0}$$ Multiply both sides by $-1$ and reverse inequality: $$x < 0$$ 7. Final solution for the first inequality is: $$\boxed{x < 0}$$