1. $\mathbf{B i}$. ** Find all functions $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ such that for any $m, n \in \mathbf{N}^{*}$, we have
$$
f^{2}(m)+f(n) \mid\left(m^{2}+n\right)^{2}
$$
An obviously satisfying function is $f(n)=n$. Below is the proof that this is the only function that meets the requirement.

Let $f(1)=t$, and set $m=n=1$, we know $t^{2}+t \mid 4$, so $t^{2}+t=2$ or 4, the latter has no solution, the former gives $f(1)=1$.

In (1), let $m=1$, we know for any $n \in \mathbf{N}^{*}$, $f(n)+1 \mid (n+1)^{2}$. Let $n=1$, we know for any $m \in \mathbf{N}^{*}$, $f^{2}(m)+1 \mid (m^{2}+1)^{2}$.

For any prime $p, f(p-1)+1 \mid p^{2} \Rightarrow f(p-1)+1=p$ or $p^{2}$. If $f(p-1)+1=p^{2}$, then $\left(p^{2}-1\right)^{2}+1 \mid \left((p-1)^{2}+1\right)^{2}$, but $\left((p-1)^{2}+1\right)^{2} \leqslant \left((p-1)^{2}+(p-1)\right)^{2}=(p-1)^{2} p^{2}$, and $\left(p^{2}-1\right)^{2}+1=p^{4}-2 p^{2}+2=p^{2}(p^{2}-2)+2 > p^{2}(p-1)^{2}$. Contradiction. Therefore, $f(p-1)+1=p$, i.e., $f(p-1)=p-1$.

Thus, there exist infinitely many $k \in \mathbf{N}^{*}$ such that $f(k)=k$. For any $n \in \mathbf{N}^{*}$, we have $f(k)^{2}+f(n) \mid (k^{2}+n)^{2}$, i.e., $k^{2}+f(n) \mid (k^{2}+n)^{2}$, and
$$
(k^{2}+n)^{2}=(k^{2}+f(n)+n-f(n))^{2}=A(k^{2}+f(n))+(n-f(n))^{2}.
$$

Where $A$ is some integer, this shows that $k^{2}+f(n) \mid (n-f(n))^{2}$. For a fixed $n$, since $(n-f(n))^{2}$ is a multiple of infinitely many positive integers, hence $(n-f(n))^{2}=0$, i.e., $f(n)=n$.