1. Definition:
$$
\begin{array}{l}
f(n)=\sqrt{n(n+1)(n+2)(n+3)+1}, \\
g(n)=(n+1)^{2} .
\end{array}
$$

Given the following conclusions:
(1) For any rational number $n, f(n)$ is a rational number;
(2) $f(1)-g(1)=1$;
(3) If $f(n)-g(n)=2012$, then $n=2012$;
(4) For any rational number $n$, $f(n)>g(n)$.

The correct conclusions are $\qquad$ (fill in the sequence numbers of all correct conclusions).
Sure, here is the translation:

```
2. (1)(2)(3).
Notice,
\[
\begin{array}{l}
\sqrt{n(n+1)(n+2)(n+3)+1} \\
=\sqrt{\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1} \\
=\left|n^{2}+3 n+1\right|,
\end{array}
\]

Thus, conclusions (1) and (2) are correct;
When \( n^{2}+3 n+1 > 0 \),
\[
f(n)-g(n)=n,
\]

In this case, \( f(n)-g(n)=2012 \), then \( n=2012 \) is correct;
\[
\begin{array}{l}
\text{When } n^{2}+3 n+1 < 0, \\
f(n)-g(n)=-2 n^{2}-5 n-2=2012 \\
\Rightarrow 2 n^{2}+5 n+2014=0,
\end{array}
\]

This equation has no solution, hence conclusion (3) is correct;
For conclusion (4), when \( n=0 \), \( f(n)=g(n) \), hence conclusion (4) is incorrect.
```