(a) Find all positive integers $n$ for which the equation $(a^a)^n = b^b$ has a solution
in positive integers $a,b$ greater than $1$.
(b) Find all positive integers $a, b$ satisfying $(a^a)^5=b^b$
### Part (a)
We need to find all positive integers \( n \) for which the equation \((a^a)^n = b^b\) has a solution in positive integers \( a, b \) greater than 1.

1. **Initial Observation**:
   - If \( n \geq 3 \), we have \((a, b) = ((n-1)^{n-1}, (n-1)^n)\).
   - If \( n = 1 \), we have \((a, b) = (3, 3)\).

2. **Case Analysis**:
   - We need to consider the cases when \( n \in \{2, 5\} \).

3. **General Form**:
   - Given \( a^{an} = b^b \), for any prime number \( p \mid b \), we have \( an \nu_p(a) = b \nu_p(b) \).
   - Hence, \( \nu_p(a) = xt \) and \( \nu_p(b) = yt \) for some positive integers \( x, y, t \) such that \( \frac{x}{y} = \frac{b}{an} \) and \( \gcd(x, y) = 1 \).
   - This implies \( a = T^x \) and \( b = T^y \) for some positive integer \( T \).
   - Therefore, \( xnT^x = yT^y \implies xn = yT^{y-x} \).

4. **Prime Factor Analysis**:
   - For any prime number \( p \), if \( p \mid y \implies p \mid n \implies p = n \).
   - So, there exists a non-negative integer \( \ell \) such that \( y = n^\ell \). It is easy to see that \( \ell \geq 1 \).
   - The equation becomes \( xT^x = n^{\ell-1}T^{n^\ell} \).

5. **Prime Divisors**:
   - If there exists a prime number \( p \mid T \) such that \( p \neq n \), we have \( \nu_p(x) + x \nu_p(T) = n^\ell \nu_p(T) \).
   - So, \( \nu_p(x) = \nu_p(T)(n^\ell - x) \).
   - Setting \( x = n^\ell - z \) where \( z \in \mathbb{Z}^+ \) in the equation gives us \( n^\ell - z = n^{\ell-1}T^z \implies n^{\ell-1} \mid z \).
   - Hence, \( \nu_p(T)n^{\ell-1} \mid \nu_p(x) \implies x \geq p^{\nu_p(T)n^{\ell-1}} \geq p^{n^{\ell-1}} \geq 2^{n^{\ell-1}} \) while \( x \leq n^\ell \).

6. **Case \( \ell = 1 \)**:
   - We have \( x \leq n \).

7. **Case \( \ell > 1 \)**:
   - We have \( n^{\ell-1} \geq 2 \). It is not hard to see that \( 2^u \geq u^2 \) for all \( u \in \mathbb{Z}_{\geq 5} \).
   - So, either \( n^{\ell-1} \in \{2, 3, 4\} \), or \( n^\ell > n^{2(\ell-1)} \implies 2 > \ell \implies \ell = 1 \), contradiction.
   - So, \( n^{\ell-1} \in \{2, 3, 4\} \implies n = 2 \) and \( \ell \in \{2, 3\} \).

8. **Otherwise**:
   - We get that \( T \) must be a power of \( n \). So \( x = \frac{n^{\ell-1}T^{n^\ell}}{T^x} \) is a power of \( n \) too.
   - But \( \gcd(x, y) = 1 \implies x = 1 \). Let \( T = n^j \). We get that \( j = \ell-1 + jn^\ell \geq jn^\ell > jd \), contradiction.

9. **Conclusion**:
   - All solutions must come from the first case.

### Part (b)
We need to find all positive integers \( a, b \) satisfying \((a^a)^5 = b^b\).

1. **Case \( n = 2 \)**:
   - Easy to see that there is no solution from the subcase \( \ell = 1 \).
   - We have \( \ell \in \{2, 3\} \). Plugging in the equation gives us \( xT^x = 2T^4 \) or \( xT^x = 4T^8 \).
   - There is no solution from the first equation.
   - From the second equation, we have \( x \leq 7 \).
   - Also, recall that \( T \) must contain a prime divisor not equal to \( 2 \).
   - So, \( x \) cannot be a power of two, which implies \( x \in \{3, 5, 6, 7\} \).
   - For any prime \( w \mid x \) such that \( w \neq 2 \), we have \( \nu_w(x) = 1 \).
   - So, \( 1 + x \nu_w(T) = 8 \nu_w(T) \implies \nu_w(T) = 1 \implies x = 7 \), which is not possible.
   - Hence, no solution in this case.

2. **Case \( n = 5 \)**:
   - We have \( \ell = 1 \implies y = 5 \) and \( xT^x = T^5 \), which gives \( x < 5 \implies x \in \{1, 2, 3, 4\} \).
   - Easy to see that the only possible case is when \( x = 4 \), which gives \( 4 = T \).
   - Hence, \((a, b) = (T^x, T^y) = (4^4, 4^5)\), which is the only solution.

The final answer is \( \boxed{ (a, b) = (4^4, 4^5) } \).