a) Note that for any integer $x$ the product $x(x+1)$ is even. This suggests the following example:
$$1008x(x+1)+2016=1008x^2+1008x+2016.$$
For every integer $x$, $f(x)$ is divisible by $2016$, but $2016$ does not divide all the coefficients ($1008$ is not divisible by $2016$). Thus, the answer is no.

b) We have $f(x) = ax^2 + bx + c$. Let us do the following substitutions:
$$\begin{aligned} x &= 0 \Rightarrow f(0) = c \implies 2017 \,|\, c; \\ x &= 1 \Rightarrow f(1) = a + b + c \implies 2017 \,|\, a + b + c; \\ x = -1 \Rightarrow f(-1) = a - b + c \implies 2017 \,|\, a - b + c. \end{aligned}$$
Then both $a + b$ and $a - b$ are divisible by $2017$, so are $2a$ and $2b$. Since $2017$ is odd, all the coefficients are divisible by $2017$.