(1) When $n = 1$, we have $a_1^2 = a_1^3$. Since $a_1 \neq 0$, we get $a_1 = 1$.
When $n = 2$, we have $(1+a_2)^2 = 1+a_2^3$. Since $a_2 \neq 0$, we get $a_2 = 2$ or $a_2 = -1$.
When $n = 3$, we have $(1+a_2+a_3)^2 = 1+a_2^3+a_3^3$. For $a_2 = 2$, we get $a_3 = 3$ or $a_3 = -2$; for $a_2 = -1$, we get $a_3 = 1$.
In summary, we get three sequences consisting of three terms that satisfy the given condition:
$$\{1, 2, 3\}, \{1, 2, -2\}, \text{ and } \{1, -1, 1\}.$$

(2) Let $S_n = a_1 + a_2 + \cdots + a_n$. Then we have
$$S_n^2 = a_1^3 + a_2^3 + \cdots + a_n^3 \ (n \in \mathbb{N}), \\ (S_n + a_{n+1})^2 = a_1^3 + a_2^3 + \cdots + a_n^3 + a_{n+1}^3.$$
Finding out the difference of the two expressions above and by $a_{n+1} \neq 0$, we have $2S_n = a_{n+1}^2 - a_{n+1}$.
When $n = 1$, we know from (1) that $a_1 = 1$.
When $n \ge 2$, we have
$$2a_n = 2(S_n - S_{n-1}) = (a_{n+1}^2 - a_{n+1}) - (a_n^2 - a_n).$$
And that is
$$(a_{n+1} + a_n)(a_{n+1} - a_n - 1) = 0.$$
Then we get $a_{n+1} = -a_n$ or $a_{n+1} = a_n + 1$.
Finally, from $a_1 = 1$ and $a_{2013} = -2012$, we find the formula of general term for a required sequence as
$$a_n = \begin{cases} n, & 1 \le n \le 2012, \\ 2012(-1)^n, & n \ge 2013. \end{cases}$$