Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.

Your way of dealing with $\pi$ is very smart, but apparently you only took it because $\pi$ is irrational, and you are expecting to get the "nicer" values of $2,4,6$ in different ways. Here's something you can do, use seperate terms that are zero at $1,2,\pi$ and $6$ at $3$. Try it for yourself. Add all the terms. You already have the requisite insight.

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{(x-\pi)(x-1)(x-2)6}{2(3-\pi)}+\frac{(x-\pi)(x-1)(x-3)4}{(2-\pi)\times-1}\\+\frac{(x-2)(x-3)(x-\pi)2}{(-1)(-2)(1-\pi)}$$

Simplyfying,

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{3(x-\pi)(x-1)(x-2)}{(3-\pi)}+\frac{4(x-\pi)(x-1)(x-3)}{(\pi-2)}+\\\frac{(x-2)(x-3)(x-\pi)2}{2(1-\pi)}$$

