Solution:

a. Since $2007$ is an odd number, the value of a polynomial at $2007$ is even or odd depending on whether the number of its odd coefficients is even or odd. But if $p$ and $q$ are similar, then in particular they contain the same number of odd coefficients, and therefore $p(2007)$ and $q(2007)$ are either both even or both odd. In either case, their difference is divisible by $2$.

b. Yes, this also holds for $k=2006$.
For every non-negative integer $h$, we have $2007^{h} \equiv 1 \pmod{2006}$. If $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ and $q(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{0}$ then we have
$$\begin{aligned} p(2007) &\equiv a_{n}+a_{n-1}+\cdots+a_{0} \quad (\bmod\ 2006) \\ q(2007) &\equiv b_{n}+b_{n-1}+\cdots+b_{0} \quad (\bmod\ 2006) \end{aligned}$$
from which $p(2007)-q(2007) \equiv (a_{n}+a_{n-1}+\cdots+a_{0})-(b_{n}+b_{n-1}+\cdots+b_{0})=0 \pmod{2006}$.

Alternative Solution:

Both cases (a) and (b) can be solved in the following way. Note that if $p$ and $q$ are similar it necessarily follows that $p(1)=q(1)$. Now let $r(x)=p(x)-q(x)$; we have $r(1)=0$, hence $(x-1)$ divides $r(x)$. But then $2006=2007-1$ divides $r(2007)=p(2007)-q(2007)$.