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Let's break down and solve the mathematical problems step by step.

### Partie I

1. **Étudier la parité de** \( A = (6n + 11)(-1)^{6n^8} + (2n + 3)(-1)^{2n+5} \)

   - Parity refers to whether the expression is even or odd. We need to examine each term to determine whether the sum is even or odd based on the powers of \(-1\).
   - \( (-1)^{6n^8} = (-1)^{\text{even}} = 1 \), because any even power of \(-1\) equals 1.
   - \( (-1)^{2n+5} = (-1)^{\text{odd}} = -1 \), since \(2n+5\) is odd.
   - So, the expression simplifies to:
     \[
     A = (6n + 11)(1) + (2n + 3)(-1) = 6n + 11 - (2n + 3) = 4n + 8
     \]
   - \( 4n + 8 \) is always even, hence **A is even**.

2. **Montrer que le nombre \( (6n + 12)^2 + (2n + 4)^2 \) est un multiple de 40.**

   - Simplifying the terms:
     \[
     (6n + 12)^2 = 36n^2 + 144n + 144
     \]
     \[
     (2n + 4)^2 = 4n^2 + 16n + 16
     \]
     Adding the two:
     \[
     (36n^2 + 144n + 144) + (4n^2 + 16n + 16) = 40n^2 + 160n + 160
     \]
     Clearly, the entire expression is divisible by 40, so it is a multiple of 40.

3. **Soit \( n \in \mathbb{N} \) tel que \( n \geq 2 \), on pose \( a = 8^{n+2} \) et \( b = 21 \times 2^4 \).**
   
   We are asked to perform two tasks here:
   
   **(a)** Décomposer en produit de facteurs premiers les nombres \( a \) et \( b \).
   - \( a = 8^{n+2} = (2^3)^{n+2} = 2^{3(n+2)} = 2^{3n+6} \)
   - \( b = 21 \times 2^4 = 3 \times 7 \times 2^4 \)

   **(b)** En déduire \( \text{PGCD}(a : b) \) et \( \text{PPCM}(a : b) \).
   - \( \text{PGCD}(a, b) \) is the greatest common divisor. The only common factor between \( a \) and \( b \) is \( 2 \), and the smallest power of 2 is \( 2^4 \), so:
     \[
     \text{PGCD}(a, b) = 2^4 = 16
     \]
   - \( \text{PPCM}(a, b) \) is the least common multiple. For the LCM, take the highest powers of all prime factors:
     \[
     \text{PPCM}(a, b) = 2^{3n+6} \times 3 \times 7
     \]

### Partie II

1. **Soit \( p \) un entier naturel impair, montrer que:**
   - **(a)** \( 8 \) divise \( p^2 - 1 \).
   - **(b)** En déduire que \( 16 \) divise \( p^4 - 1 \).

   For \( p \) odd, we know \( p^2 - 1 = (p - 1)(p + 1) \). Both \( p-1 \) and \( p+1 \) are even, and one of them is divisible by 4. Therefore, \( p^2 - 1 \) is divisible by 8.

   To show that \( p^4 - 1 \) is divisible by 16, factor \( p^4 - 1 = (p^2 - 1)(p^2 + 1) \). Since \( p^2 - 1 \) is divisible by 8 and \( p^2 + 1 \) is even, the product is divisible by 16.

2. **Simplifier les expressions suivantes:**
   - **\( A = 2 + \frac{4}{\frac{4}{5} + \frac{1}{2}} \)**

   First, simplify the denominator:
   \[
   \frac{4}{5} + \frac{1}{2} = \frac{8}{10} + \frac{5}{10} = \frac{13}{10}
   \]
   So, the expression becomes:
   \[
   A = 2 + \frac{4}{\frac{13}{10}} = 2 + 4 \times \frac{10}{13} = 2 + \frac{40}{13} = \frac{26}{13} + \frac{40}{13} = \frac{66}{13}
   \]

This concludes the main solutions. Would you like me to go into more detail on any part, or tackle other questions? Here are some additional related problems:

1. What is the method to simplify expressions involving multiple fractions?
2. How do you factor quadratic expressions effectively?
3. What is the significance of finding the greatest common divisor (GCD) and least common multiple (LCM)?
4. How can the parity of an expression help in determining divisibility properties?
5. What are key properties of powers of integers, especially odd and even numbers?

**Tip:** When simplifying algebraic expressions, it's often helpful to simplify the denominator first when fractions are involved!

Study the parity of A = (6n + 11)(-1)^{6n^8} + (2n + 3)(-1)^{2n+5} and solve various algebraic expressions and factorization problems.

Explore detailed solutions for algebraic problems on parity, prime factorization, GCD, and LCM. These problems involve simplifying complex expressions and factorizing integers using prime factors. Suitable for Grade 12 and above.

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