Solution:

a. Yes, very soon, in fact the next four terms (from $7$ to $10$th) are $2, 0, 0, 7$.

b. We will prove that $1, 8, 4, 0$ will be a subsequence again. Assume the contrary. Since there are only finitely many combinations of four digits (precisely $10^4$), and the sequence is infinite, some combination of four digits (say $(a, b, c, d)$) has to reappear. Assume that $(x_n, x_{n+1}, x_{n+2}, x_{n+3}) = (a, b, c, d)$ is the first occurrence of $(a, b, c, d)$ and that $(x_m, x_{m+1}, x_{m+2}, x_{m+3}) = (a, b, c, d)$ is the second. Clearly $m > n$. However, $x_{n-1}$ and $x_{m-1}$ are uniquely determined and they have to be the same numbers. Thus $x_{n-1} = x_{m-1}$, $x_{n-2} = x_{m-2}$, and so on. This means that $x_1 = x_{m-n+1}$, $x_2 = x_{m-n+2}$, $x_3 = x_{m-n+3}$, and $x_4 = x_{m-n+4}$, which is a contradiction.