1. Can we obtain the triplet of numbers $2,6,9$ in some order from the triplet with numbers $2,4,7$?
Solution:

If $f$ is a polynomial with integer coefficients, then $f(x)-f(y)$ is divisible by $x-y$ for any integers $x, y$. Now notice that $7-2=5$, and among the numbers $2, 6, 9$ there are no such numbers whose difference is divisible by 5. (Here, either apply the above statement to the polynomial that is the composition of the applied polynomials, or simply notice that in the sequence $x-y, f(x)-f(y), g(f(x))-g(f(y)), \ldots$, where $f, g, \ldots$ are the applied trinomials, each term is divisible by the previous one, and thus by the very first difference). Answer: no.