We are given the equation:

$a + \frac{1}{a+2} = 3$

Step 1: Solve for $a$

To isolate $a$, subtract $\frac{1}{a+2}$ from both sides of the equation:

$a = 3 - \frac{1}{a+2}$

Next, multiply both sides of the equation by $a+2$ to eliminate the denominator on the right side:

$a(a+2) = (3(a+2)) - 1$

Simplifying this equation:

$a^2 + 2a = 3a + 6 - 1$

$a^2 + 2a = 3a + 5$

Subtract $3a$ from both sides:

$a^2 - a = 5$

Rearrange into a standard quadratic form:

$a^2 - a - 5 = 0$

Step 2: Solve the quadratic equation

We can solve $a^2 - a - 5 = 0$ using the quadratic formula:

$a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)}$

$a = \frac{1 \pm \sqrt{1 + 20}}{2}$

$a = \frac{1 \pm \sqrt{21}}{2}$

Thus, the two possible values for $a$ are:

$a = \frac{1 + \sqrt{21}}{2} \quad \text{or} \quad a = \frac{1 - \sqrt{21}}{2}$

Step 3: Solve $(a+2)^5 - \frac{1}{(a+2)^5}$

Now, substitute $a$ into $(a+2)^5 - \frac{1}{(a+2)^5}$.

Since we know from the original equation that:

$a + \frac{1}{a+2} = 3$

Multiply both sides by $a+2$:

$(a+2) + \frac{1}{a+2} = 3$

This implies:

$a + 2 = 3$

Thus, $a+2 = 3$.

Step 4: Calculate $(a+2)^5 - \frac{1}{(a+2)^5}$

Using $a+2 = 3$:

$(3)^5 - \frac{1}{(3)^5} = 243 - \frac{1}{243}$

This simplifies to:

$243 - \frac{1}{243} = \frac{243^2 - 1}{243} = \frac{59049 - 1}{243} = \frac{59048}{243}$

Therefore, the solution is:

$\frac{59048}{243}$

Would you like more details on any step? Here are some related questions:

1. How does the quadratic formula work in detail?
2. Can we solve this problem by a different method?
3. What are some properties of exponents used in the final step?
4. How do we simplify fractions with large numbers?
5. Can this type of equation be solved graphically?

Tip: Always check if your solution satisfies the original equation by substituting back into it.