Do not use a calculator in answering this question. Solve the inequality
$\frac{x+4}{-x^2 + 2x + 3} < 1.$

1. Determine the solution to the main inequality (from the previous part).

From the original inequality $\frac{x+4}{-x^2 + 2x + 3} < 1$, we rearrange to get:
$\frac{x+4 - (-x^2+2x+3)}{-x^2+2x+3} < 0$
$\frac{x^2-x+1}{-(x-3)(x+1)} < 0$
Since the discriminant of $x^2-x+1$ is $(-1)^2 - 4(1)(1) = -3 < 0$ and the coefficient of $x^2$ is positive, $x^2-x+1 > 0$ for all real $x$.
Therefore, $-(x-3)(x+1) < 0 \implies (x-3)(x+1) > 0$.
The solution to the main inequality is $x < -1 \text{ or } x > 3$.

2. Apply the substitution $x \to x^2$ to deduce the new inequality's conditions.

The inequality $\frac{x^2+4}{-x^4 + 2x^2 + 3} < 1$ is identical to the main inequality with $x$ replaced by $x^2$.
Substituting $x^2$ into the solution gives:
$x^2 < -1 \quad \text{or} \quad x^2 > 3$
Since $x^2 \ge 0$ for all real $x$, $x^2 < -1$ has no real solutions.

3. Solve the remaining valid inequality.

We solve $x^2 > 3$:
$x^2 - 3 > 0$
$(x - \sqrt{3})(x + \sqrt{3}) > 0$
$x < -\sqrt{3} \quad \text{or} \quad x > \sqrt{3}$

1. Solve the main inequality from the first part to establish the base solution.

$\frac{x+4}{-x^2 + 2x + 3} - 1 < 0$
$\frac{x+4 - (-x^2 + 2x + 3)}{-x^2 + 2x + 3} < 0$
$\frac{x^2 - x + 1}{-(x^2 - 2x - 3)} < 0$
Since $x^2 - x + 1 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4} > 0$ for all real $x$, the inequality holds when the denominator is negative:
$-(x - 3)(x + 1) < 0$
$(x - 3)(x + 1) > 0$
This gives the base solution: $x < -1 \text{ or } x > 3$.

2. Relate the target inequality to the original inequality by making a substitution.

Substitute $x$ with $-x$ into the original inequality $\frac{x+4}{-x^2 + 2x + 3} < 1$:
$\frac{(-x)+4}{-(-x)^2 + 2(-x) + 3} < 1$
$\frac{-x+4}{-x^2 - 2x + 3} < 1$
Multiply the numerator and denominator of the fraction by $-1$:
$\frac{x-4}{x^2 + 2x - 3} < 1$
This matches the inequality given in part (ii).

3. Deduce the final solution by applying the same substitution to the base solution.

Since the target inequality is formed by replacing $x$ with $-x$ in the original inequality, we replace $x$ with $-x$ in the base solution $x < -1 \text{ or } x > 3$:
$-x < -1 \quad \text{or} \quad -x > 3$
Multiplying by $-1$ and flipping the inequality signs gives:
$x > 1 \quad \text{or} \quad x < -3$