Find all positive integers $n$ such that the number $A_n =\frac{ 2^{4n+2}+1}{65}$ is
a) an integer,
b) a prime.
Let's solve the problem step-by-step.

### Part (a): Finding when \( A_n \) is an integer

1. **Expression Simplification:**
   \[
   A_n = \frac{2^{4n+2} + 1}{65}
   \]
   For \( A_n \) to be an integer, the numerator \( 2^{4n+2} + 1 \) must be divisible by 65. Since \( 65 = 5 \times 13 \), we need \( 2^{4n+2} + 1 \) to be divisible by both 5 and 13.

2. **Divisibility by 5:**
   \[
   2^{4n+2} + 1 \equiv 0 \pmod{5}
   \]
   We know that \( 2^4 \equiv 1 \pmod{5} \), so:
   \[
   2^{4n+2} = 2^{4n} \cdot 2^2 \equiv 1^n \cdot 4 \equiv 4 \pmod{5}
   \]
   Thus:
   \[
   2^{4n+2} + 1 \equiv 4 + 1 \equiv 0 \pmod{5}
   \]
   This condition is always satisfied.

3. **Divisibility by 13:**
   \[
   2^{4n+2} + 1 \equiv 0 \pmod{13}
   \]
   We know that \( 2^{12} \equiv 1 \pmod{13} \) (by Fermat's Little Theorem), so:
   \[
   2^{4n+2} = 2^{4n} \cdot 2^2 \equiv (2^4)^n \cdot 4 \equiv 3^n \cdot 4 \pmod{13}
   \]
   Thus:
   \[
   2^{4n+2} + 1 \equiv 4 \cdot 3^n + 1 \equiv 0 \pmod{13}
   \]
   This simplifies to:
   \[
   4 \cdot 3^n \equiv -1 \pmod{13}
   \]
   Multiplying both sides by the modular inverse of 4 modulo 13, which is 10 (since \( 4 \cdot 10 \equiv 1 \pmod{13} \)):
   \[
   3^n \equiv 10 \cdot (-1) \equiv -10 \equiv 3 \pmod{13}
   \]
   Therefore:
   \[
   3^n \equiv 3 \pmod{13}
   \]
   This implies:
   \[
   n \equiv 1 \pmod{3}
   \]
   So \( n = 3k + 1 \) for some integer \( k \).

### Part (b): Checking if \( A_n \) is a prime

1. **Expression for \( A_n \):**
   \[
   A_n = \frac{2^{4n+2} + 1}{65}
   \]
   Substituting \( n = 3k + 1 \):
   \[
   A_n = \frac{2^{4(3k+1)+2} + 1}{65} = \frac{2^{12k+6} + 1}{65}
   \]

2. **Factorization:**
   \[
   2^{12k+6} + 1 = (2^{6k+3} - 2^{3k+1} + 1)(2^{6k+3} + 2^{3k+1} + 1)
   \]
   We need to check if either factor can be 65 or a prime number.

3. **Checking for \( k \geq 1 \):**
   For \( k \geq 1 \):
   \[
   2^{6k+3} - 2^{3k+1} + 1 \geq 497
   \]
   This implies:
   \[
   2^{6k+3} + 2^{3k+1} + 1 \leq 65
   \]
   This is a contradiction because the product of two numbers, one of which is at least 497, cannot be 65 times a prime number.

4. **Checking for \( k = 0 \):**
   For \( k = 0 \):
   \[
   n = 1 \quad \text{and} \quad A_1 = \frac{2^{4 \cdot 1 + 2} + 1}{65} = \frac{2^6 + 1}{65} = \frac{65}{65} = 1
   \]
   Since 1 is not a prime number, \( A_n \) is not prime for \( n = 1 \).

Thus, \( A_n \) is not prime for any \( n \).

The final answer is \( \boxed{ n = 3k + 1 } \) for part (a) and \( A_n \) is not prime for any \( n \) for part (b).