By expressing $\frac{3x - 7}{(x + 3)(x - 2)} - 1$ as a single simplified fraction, solve exactly the inequality $\frac{3x - 7}{(x + 3)(x - 2)} < 1.$ Hence, solve exactly the inequality $\frac{3e^x - 7}{(e^x + 3)(e^x - 2)} < 1$.

1. Express the given expression as a single simplified fraction.

$\begin{aligned} \frac{3x - 7}{(x + 3)(x - 2)} - 1 &= \frac{3x - 7 - (x + 3)(x - 2)}{(x + 3)(x - 2)} \\ &= \frac{3x - 7 - (x^2 + x - 6)}{(x + 3)(x - 2)} \\ &= \frac{-x^2 + 2x - 1}{(x + 3)(x - 2)} \\ &= \frac{-(x - 1)^2}{(x + 3)(x - 2)} \end{aligned}$

2. Set up the rational inequality and determine the required sign of the denominator.

We are solving the inequality $\frac{-(x - 1)^2}{(x + 3)(x - 2)} < 0$.
Since the square of any real number is non-negative, $(x - 1)^2 \ge 0$, which means the numerator $-(x - 1)^2 \le 0$.
For the entire fraction to be strictly less than $0$, the numerator must be non-zero (so $x \neq 1$) and the denominator must be strictly positive:
$(x + 3)(x - 2) > 0$

3. Solve the quadratic inequality for $x$.

The critical values for the denominator are $x = -3$ and $x = 2$.
The parabola $y = (x + 3)(x - 2)$ opens upwards, so it takes positive values outside the roots:
$x < -3 \quad \text{or} \quad x > 2$
Note that $x=1$ does not fall in this range, so the condition $x \neq 1$ is naturally satisfied.

1. Determine the solution to the inequality from the previous part.

From the previous part, the inequality can be rewritten as:
$\frac{3x - 7}{(x + 3)(x - 2)} - 1 < 0$
$\frac{-(x - 1)^2}{(x + 3)(x - 2)} < 0$
Since $-(x - 1)^2 \le 0$ for all real $x$ and equals $0$ when $x = 1$, the numerator is strictly negative for $x \ne 1$.
For the entire fraction to be negative, the denominator must be positive:
$(x + 3)(x - 2) > 0 \quad \text{and} \quad x \ne 1$
Solving this gives:
$x < -3 \quad \text{or} \quad x > 2$

2. Substitute $e^x$ into the established solution.

The new inequality is identical in form to the original, with $x$ replaced by $e^x$. Substituting $x = e^x$ into the solution bounds gives:
$e^x < -3 \quad \text{or} \quad e^x > 2$

3. Solve for $x$, rejecting any invalid cases.

Since $e^x > 0$ for all real $x$, the inequality $e^x < -3$ has no real solutions.
We are left with:
$e^x > 2$
Taking the natural logarithm on both sides yields:
$x > \ln 2$