a) Take $q=0$ then we need $x=1^{a_1}2^{a_2}\cdots (n+1)^{a_n}$. Then it's easy: we write down factors $p^a$ such that the $p$ are distinct primes and we pick the exponents such that the product is $x$. b) No, take $q=-6/7$ then $x=1$ is char. Now suppose that 2 is char, hence $2=({1\over 7})^{a_1}\cdots ({{7k-6}\over 7})^{a_k}$. Let $b_i:=|a_i|$ then $2=({{\alpha_1}\over {\beta_1}})^{b_1}\cdots ({{\alpha_k}\over {\beta_k}})^{b_k}$. Here $\{\alpha_i,\beta_i\}=\{7,7i-6\}$. The factors 7 should cancel, hence have $2={p\over q}$ where $p,q\in Z, p,q\equiv 1(7)$. Contradiction.