1)Since $b-a=(p(T)-a)-(p(T)-b)=2k+1-(2l+1)=2(k-l)$ it follows that all elements differ by an even integer. So if $a$ satisfies the property so do all other elements. Therefore assume the set is of the form $\{x, x+2, x+4,...,x+2\cdot 2020\}$. Note that, since the degree is odd, the equation $x(x+2)\cdots (x+4040)=x+1$ has at least one root. By rational root theorem if it were relational it would have to be $\pm 1$ but by size reasons it is easy to see that this is impossible and thus there exists a irrational root $x_0$. Therefore the set $\{x_0,x_0+2,...,x_0+4040\}$ satisfies the condition. 2) By the same logic as before, they must all be rational numbers which differ by an even integer. Since translating by an integer doesn't change the denominator (the new numerators will always be coprime with the denominator, since the old numerator was as well), it follows that if $v_p(x)=-a<0$, then $v_p(p(T))=-2021a$. However this must als be the same of $v_p(x+2k+1)=-a$, and so $a=0$, ie. the rationals have no denominator and are integers. Now $\pmod 2$ gives a contradiction since we must either have $p(T)-a\equiv 1\cdot 1\cdots 1-1\equiv 0$ or $0\cdot 0\cdots 0-0\equiv 0$ since all integers differ by an even amount. Therefore it is impossible to have at least one rationa number.