Solve the inequality | 3x + 2 < 4x + 5.

$A < B$ and $A > -B$. Additionally, for a solution to exist, $B$ must be positive, so $B > 0$.

Condition 1: $3x + 2 < 4x + 5$

$$2 - 5 < 4x - 3x$$ $$-3 < x$$ $$x > -3$$

Condition 2: $3x + 2 > -(4x + 5)$

$$3x + 2 > -4x - 5$$ $$3x + 4x > -5 - 2$$ $$7x > -7$$ $$x > -1$$

Condition 3: $4x + 5 > 0$

$$4x > -5$$ $$x > -\frac{5}{4}$$ $$x > -1.25$$

We need $x > -3$, $x > -1$, and $x > -1.25$.
The most restrictive condition that satisfies all three is $x > -1$.

The solution to the inequality is $\boxed{x > -1}$.

2. Find the coefficient of $x^3$ in the expansion of $(3x - \frac{2}{x})^7$.

The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
In this case, $a = 3x$, $b = -\frac{2}{x} = -2x^{-1}$, and $n = 7$.

$$T_{r+1} = \binom{7}{r} (3x)^{7-r} (-2x^{-1})^r$$

$$T_{r+1} = \binom{7}{r} 3^{7-r} x^{7-r} (-2)^r (x^{-1})^r$$ $$T_{r+1} = \binom{7}{r} 3^{7-r} (-2)^r x^{7-r} x^{-r}$$ $$T_{r+1} = \binom{7}{r} 3^{7-r} (-2)^r x^{7-2r}$$

We want the coefficient of $x^3$, so:

$$7 - 2r = 3$$ $$4 = 2r$$ $$r = 2$$

The coefficient is $\binom{7}{2} 3^{7-2} (-2)^2$.

$$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21$$ $$3^{7-2} = 3^5 = 243$$ $$(-2)^2 = 4$$

$$\text{Coefficient} = 21 \times 243 \times 4$$ $$\text{Coefficient} = 21 \times 972$$ $$\text{Coefficient} = 20412$$

The coefficient of $x^3$ is $\boxed{20412}$.