Some properties of zero-sets and cozero-sets

Some ideas for 2: If $F,G$ are zero-sets (as is their normal name in English) we an write $F=f^{-1}[\{0\}$ and $G=g^{-1}[\{0\}$ for some continuous $f,g:X \to \Bbb R$

But then $F \cap G = h^{-1}[\{0\}]$ where $h(x)=f^2(x) + g^2(x)$ is also continuous.

And $F \cup G = (fg)^{-1}[\{0\}]$ and $fg$ is also continuous.

If $X$ is completely regular, $U$ is open and $x \in U$, we can find a continuous $f: X \to [0,1]$ with $f(x)=0$ and $f[X\setminus U]=\{1\}$. Then $V:=f^{-1}[[0,1)]$ is a cozero-set and $x \in V \subseteq U$, so cozero sets are a base.

The converse is quite similar.

For part 1, your argument is correct but it does not prove what you were asked to prove. If you define $f(x) = 1$ for all $x$, then $f^{-1}(\{0\}) = \emptyset$, which shows that $\emptyset$ is null and that $X$ is conull. (Your argument shows that $\emptyset$ is conull and $X$ is null, but that can be true vice versa too.)

What you are being asked to prove in part 2 was false (as originally stated without any restrictions one the unions and intersections). If $X= \Bbb{R}$, then any singleton set $\{x\}$ is null, so, if part 2 were true, any set would be null, since it is the union of its singleton subsets, but the only non-empty open subset of $\Bbb{R}$ that is null is $\Bbb{R}$ itself (because $f^{-1}(\{0\})$ is closed, and $\emptyset$ and $\Bbb{R}$ are the only subsets of $\Bbb{R}$ that are both open and closed). E.g., $(0, 1)$ is a subset of $\Bbb{R}$ that is not null.

I defer to Henno's answer to the now corrected part 2 and to parts 3 and 4.