a.
Recall the fact that $b - c \mid f(b) - f(c)$ for any integers $b$ and $c$. Putting $b = a_{i+1}$ and $c = a_i$, we immediately obtain
$$a_{i+1} - a_i \mid f(a_{i+1}) - f(a_i) = a_{i+2} - a_{i+1}.$$
By induction on $j$, $a_{i+1} - a_i \mid a_{j+1} - a_j$ for any $j > i$.

b.
Suppose on the contrary that $a_{2008} = 0$. Note that $a_{2009}-a_{2008} = f(0) = a_2-a_1$. By part (a), since $a_2 - a_1 \mid a_{k+1} - a_k \mid a_{2009} - a_{2008}$ for $k = 1, 2, \dots, 2007$, each $a_{k+1} - a_k$ must be $\pm f(0)$. Now, we have
$$0 = a_{2008} - a_1 = \sum_{k=1}^{2007} (a_{k+1} - a_k) = m f(0),$$
where $m \in \{-2007, -2005, \dots, -1, 1, 3, \dots, 2007\}$. This implies $f(0) = 0$, which contradicts the assumption that $c_0 \neq 0$. Therefore, $a_{2008} \neq 0$.