We are analyzing the piecewise function \( f(x) \) defined as follows:

\[
f(x) =
\begin{cases}
\frac{x^2 - 4}{x^2 + 6x + 8}, & \text{if } x < -2 \\
3x + 4, & \text{if } -2 \leq x \leq 2 \\
\frac{\sqrt{x+2} - 2}{x-2}, & \text{if } x > 2
\end{cases}
\]

To evaluate the function at specific points \( x = -2 \) and \( x = 2 \), and to simplify each piece, we proceed step by step:

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### **Step 1: Simplify the first case (\( x^2 - 4)/(x^2 + 6x + 8) \)) for \( x < -2 \):**

- Factorize numerator and denominator:
\[
x^2 - 4 = (x-2)(x+2), \quad x^2 + 6x + 8 = (x+4)(x+2)
\]

- Cancel the common factor \( (x+2) \), valid as \( x \neq -2 \):
\[
\frac{x^2 - 4}{x^2 + 6x + 8} = \frac{x-2}{x+4}.
\]

For \( x < -2 \), \( f(x) = \frac{x-2}{x+4} \).

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### **Step 2: Evaluate \( f(x) \) at \( x = -2 \):**

For \( x = -2 \), use the second piece of the function (\( 3x + 4 \)), since the condition \( -2 \leq x \leq 2 \) includes \( x = -2 \):
\[
f(-2) = 3(-2) + 4 = -6 + 4 = -2.
\]

---

### **Step 3: Simplify \( \frac{\sqrt{x+2} - 2}{x-2} \) for \( x > 2 \):**

This is an indeterminate form \( \frac{0}{0} \) when \( x = 2 \), so we rationalize the numerator:

1. Multiply numerator and denominator by the conjugate:
\[
\frac{\sqrt{x+2} - 2}{x-2} \cdot \frac{\sqrt{x+2} + 2}{\sqrt{x+2} + 2} = \frac{(x+2) - 4}{(x-2)(\sqrt{x+2} + 2)} = \frac{x-2}{(x-2)(\sqrt{x+2} + 2)}.
\]

2. Cancel the common factor \( (x-2) \), valid for \( x > 2 \):
\[
\frac{\sqrt{x+2} - 2}{x-2} = \frac{1}{\sqrt{x+2} + 2}, \quad x > 2.
\]

Thus, for \( x > 2 \), \( f(x) = \frac{1}{\sqrt{x+2} + 2} \).

---

### **Step 4: Evaluate \( f(x) \) at \( x = 2 \):**

For \( x = 2 \), use the second piece of the function (\( 3x + 4 \)), since \( -2 \leq x \leq 2 \) includes \( x = 2 \):
\[
f(2) = 3(2) + 4 = 6 + 4 = 10.
\]

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### **Final Results:**
1. For \( x < -2 \): \( f(x) = \frac{x-2}{x+4} \).
2. For \( -2 \leq x \leq 2 \): \( f(x) = 3x + 4 \).
3. For \( x > 2 \): \( f(x) = \frac{1}{\sqrt{x+2} + 2} \).
4. \( f(-2) = -2 \) and \( f(2) = 10 \).