Assume $f:\mathbb N_0\to\mathbb N_0$ is a function such that $f(1)>0$ and, for any nonnegative integers $m$ and $n$,
$$f\left(m^2+n^2\right)=f(m)^2+f(n)^2.$$(a) Calculate $f(k)$ for $0\le k\le12$.
(b) Calculate $f(n)$ for any natural number $n$.
1. **Define the set \( S \):**
   We define the set \( S := \{ n \in \mathbb{N}_0 \mid f(n) = n \} \).

2. **Initial elements in \( S \):**
   It is given that \( f(1) > 0 \). We can also see that \( f(0) = 0 \) because:
   \[
   f(0^2 + 0^2) = f(0) = f(0)^2 + f(0)^2 \implies f(0) = 0.
   \]
   Therefore, \( 0 \in S \).

3. **Prove \( 1 \in S \):**
   Since \( f(1) > 0 \) and \( f(1^2 + 0^2) = f(1) = f(1)^2 + f(0)^2 \implies f(1) = f(1)^2 \). Given \( f(1) > 0 \), the only solution is \( f(1) = 1 \). Therefore, \( 1 \in S \).

4. **Inductive step:**
   Assume \( m, n \in S \) implies \( f(m) = m \) and \( f(n) = n \). We need to show that \( f(m^2 + n^2) = m^2 + n^2 \).

5. **Using the functional equation:**
   Given \( f(m^2 + n^2) = f(m)^2 + f(n)^2 \), if \( m, n \in S \), then:
   \[
   f(m^2 + n^2) = m^2 + n^2.
   \]
   This shows that if \( m, n \in S \), then \( m^2 + n^2 \in S \).

6. **Prove \( S = \mathbb{N}_0 \):**
   We will use induction to show that \( S = \mathbb{N}_0 \).

   - **Base case:** We have already shown that \( 0, 1 \in S \).

   - **Inductive hypothesis:** Assume \( k \in S \) for all \( k \leq n \).

   - **Inductive step:** We need to show that \( n+1 \in S \).

     - If \( n+1 \) is odd, let \( n+1 = 2m + 1 \). We know \( 1, 3, 5 \in S \). Consider:
       \[
       (n+1)^2 + (m-2)^2 = (m+2)^2 + (2m-1)^2.
       \]
       Since \( m-2, m+2, 2m-1 < n+1 \), by the inductive hypothesis, they are in \( S \). Therefore, \( n+1 \in S \).

     - If \( n+1 \) is even, let \( n+1 = 2m \). We know \( 2, 4, 6 \in S \). Consider:
       \[
       (n+1)^2 + (m-5)^2 = (2m-4)^2 + (m+3)^2.
       \]
       Since \( m-5, 2m-4, m+3 < n+1 \), by the inductive hypothesis, they are in \( S \). Therefore, \( n+1 \in S \).

7. **Conclusion:**
   By induction, \( S = \mathbb{N}_0 \). Therefore, \( f(n) = n \) for all \( n \in \mathbb{N}_0 \).

\(\blacksquare\)

The final answer is \( \boxed{ f(n) = n } \).