Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to find a positive integer $n$ such that $f(n)=3$.
(1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.

If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, 3 n+1)=1$, we have $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Therefore, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$. This is a contradiction.
Hence, $f(n) \neq 2$.
(2) When $n=8$, since $3 n^{2}+n+1=201$, thus, $f(8)=3$.