a)Clearly $|M_0|=1$.Now we will prove that $|M_{2k+1}|=0,\forall k\in\mathbb{Z}$.Suppose for the sake of contradiction that $|M_{2k+1}|>0$.Let $x\in\mathbb{Q}$ be a rational number such that $x^3-2015x=2k+1$ and let $x=\frac{p}{r},p,r\in\mathbb{Z},(p,r)=1,r>0$.We have $\frac{p^3}{q^2}-2015p=q(2k+1)\in\mathbb{Z}$,so $\frac{p^3}{q^2}\in\mathbb{Z}$,but $p\neq 0$(because $x\neq 0$),implying $q^2|p^3$.But $(q,p)=1$,so $q|1$ which implies $q=1$,meaning that $x\in\mathbb{Z}$.Thus $2k+1=x^3-2015x\equiv x-x\equiv 0(mod 2)$,contradiction!Therefore $|M_{2k+1}|=0,\forall k\in\mathbb{Z}$. b)We will prove that $|M_q|\in\{0,1\}$.Suppose that there is some rational number $q$ such that $|M_q|\ge 2$.Let $x,y\in\mathbb{Q},x\neq y$ be rational numbers such that $x^3-2015x=y^3-2015y=q$.We have $(x^3-2015x)-(y^3-2015y)=(x-y)(x^2+xy+y^2-2015)=0$ and since $x\neq y$ we obatin $x^2+xy+y^2=2015$.Let $x=\frac{p}{t},y=\frac{y}{t},p,y,t\in\mathbb{Z},t\neq 0$.Then we have $p^2+pr+r^2=2015t^2$,so $5|p^2+pr+r^2|(p-r)^2$,which implies $p\equiv r(mod 5)$,so $0\equiv p^2+pr+r^2\equiv 3p^2(mod 5)$,resulting that $p\equiv r\equiv 0(mod 5)$.Let $p=5p_1,r=5p_1$.$25p_1^2+25p_1r_1+25r_1^2=2015t^2$,so $25|2015t^2$,which means $5|t^2$ i.e. $5|t$.Let $t=5t_1$.Then $p_1^2+p_1r_1+r_1^2=2015t_1^2$.By the same method we obtain $5|p_1,5|r_1,5|t_1$ and so on,so we will obtain $5^n|p,5^n|r,5^n|t,\forall n\in\mathbb{N}$,which is equivalent to $p=r=t=0$,but $t\neq 0$,contradiction! Therefore,$|M_q|<2$ and by using a) we obtain $|M_q|\in\{0,1\}$.