The zero entries in the character table of a finite group

A partial answer to Question 2: the following is a theorem of Burnside (see e.g. Isaacs, Theorem 3.8).

Theorem. Let $\chi$ be an irreducible character, let $K$ be a conjugacy class of $G$, and let $g\in K$. Suppose that $\gcd(\chi(1),\#K)=1$. Then either $\chi(g)=0$ or the irreducible representation with character $\chi$ sends $g$ to a scalar (i.e. $g \in {\rm Z}(\chi)$).

The proof uses the integrality property of central characters that John Murray mentions.

Since $\#K=\#G/\#{\rm C}_G(g)$, the hypothesis translates to the assumption that for every prime $p$ that divides $\chi(1)$, the centraliser ${\rm C}_G(g)$ contain a Sylow $p$-subgroup of $G$, so it is a stronger variant of the hypothesis in your Question 2. If for all $p|\chi(1)$ the Sylow $p$-subgroups of $G$ have order $p$, then the hypothesis of Burnside and of your question coincide.

Here are infinitely many examples showing that the answer to question 1 is negative. Take $n \equiv 1 \bmod 4$ with $n>5$ and let $\chi$ be the character of the symmetric group $S_n$ associated to the partition $(n-2,2)$. Then, as is well-known, for each $w \in S_n$, $\chi(w)$ is obtained by subtracting the number of fixed points of $w$ from the number of $2$-sets fixed (setwise) by $w$. In particular, $\chi(1)={{n} \choose {2}}-n=\frac{n(n-3)}{2}$, and if $w$ has cycle type $(n-4,4)$ then $\chi(w)=0$. Now $w$ generates its own centralizer, which thus has order $4(n-4)$, which is manifestly coprime with $\frac{n(n-3)}{2}$ under the given conditions.