The question is worded in a tricky way. When it says "polynomial" it really means "all functions that represent a polynomial complexity class". Essentially, they want the asymptotic notation for the functions $f(n)=1$, $f(n)=n$, $f(n)=n^2$, etc. The reason we want this is because you would see this inside big-O like $O(1)$, $O(n)$, $O(n^2)$, etc. Those functions represent these big-O sets.
The reason we talk about it this way is because we are talking about it in the world of complexity classes. An example is $P$.
\begin{align} P = DTIME(poly(n)) = \bigcup\limits_{k\in\mathbb{N}}DTIME(n^k) \end{align}
From the wiki on DTIME.
If a problem of input size $n$ can be solved in $O(f(n))$, we have a complexity class $DTIME(f(n))$ (or $TIME(f(n))$).
So the asymptotic notation for the set of these functions is $n^{O(1)}$. The following is a way to reason about it. \begin{align} n^{O(1)} &= n^{\{0,1,2,...\}} \\ &= \{1,\ n,\ n^2,\ ...\} \end{align}
To show that $2^{O(\log n)} = poly(n)$ is with the same reasoning as above. \begin{align} 2^{O(\log{n})} &= 2^{\{0\cdot\log{n},\ 1\cdot\log{n},\ 2\cdot\log{n},\ ...\}} \\ &= 2^{\{0,\ \log(n^1),\ \log(n^2),\ ...\}} \\ &= \{2^0,\ 2^{\log(n)},\ 2^{\log(n^2)},\ ...\} \\ &= \{1,\ n^1,\ n^2,\ ...\} \\ &= n^{O(1)} \end{align}
To formally show these are equal, you would have to use set notation.
Finally, "polynomial in x" means that $poly(x)$ grows polynomially depending on $x$. You can see that some of the entries in the table have a function as the argument to $poly$. For instance, $poly(\log{n})$ (polylogarithmic) grows slower than $poly(n)$ (polynomial) since $O(\log{n}) < O(n)$.
