Some properties of zero-sets and cozero-sets 
Definition
A subset $Y$ of a topological space is zero-set if there exist a continuous real function $f:X\rightarrow\Bbb{R}$ such that $Y=f^{-1}(\{0\})$ and so we say that a subset $Z$ of $X$ is cozero-set if $X\setminus Z$ is zero-set.
So now we prove the following statements:
  
  
*
  
*any zero-set is closed and any conul set is open and moreover the sets $\varnothing$ and $X$ are respectively zero-set and cozero-set;
  
*the finite union and the finite intersection of zero-sets or cozero-sets is respectively zero-set or cozero-set;
  
*a $T_1$ space $X$ is completely regular iff for any $x\in X$ the collection of cozero-set neighborhoods of $x$ is a local basis for $x$;
  
*if $X$ is completely regular then the collection of cozero open sets is a basis.
  
  
  Proof. Since in the above definition $f$ is a continuous function and since $\{0\}$ is closed in the euclidean topology -indeed $(\Bbb{R},\mathcal{T}_e)$ is a $T_1$ space- it result that $f^{-1}(\{0\})$ is closed and so a zero-set is closed and thus a cozero-set is open. Then since the costant function $f:X\owns x\rightarrow 0\in\Bbb{R}$ is trivially continuous and since $f^{-1}(\{0\})=X$ it result that $X$ is zero-set and $\varnothing$ is cozero-set. So we proved 1.
Now we suppose that $X$ a comletely regular space and so let be $x\in U$ a open neighborhood of $x$: since $X$ is a Tychonoff space we can claim that there exist a continuous function $f:X\rightarrow[0,1]$ such that $f(X\setminus U)=\{0\}$ and so this means that every open sets are cozero-set and so the local basis of open neighborhood of $x$ is a local basis of cozero-sets. Therefore now we suppose that $X$ is a $T_1$ space such that any its point have a cozero-set local basis and so let be $F$ a closed set and $x\notin F$: since $X\setminus F$ is an open neighborhood of $x$ there exist a cozero-set neighborhood $U$ of $x$ such $x\in U\subseteq X\setminus F$, from which it result $F\subseteq X\setminus U$; and since $U$ is cozero-set there exist a coutinuous function $f:X\rightarrow\Bbb{R}$  such that $(X\setminus U)=f^{-1}(\{0\})$ and so it is clear that $f(F)=(\{0\})$ and $f(x)\neq 0$ and so for what I proved here -is it correct?- we can claim that $X$ is a completely regular space. So we proved 3.
Now we suppose that $X$ is completely regular and so for what we prove above we know that for any $x\in X$ the collection of open cozero-set neighborhoods $\mathcal{B}(x)$ of $x$ is a local basis for $x$: so it is clear that the collection of $\mathcal{B}:=\bigcup_{x\in X}\mathcal{B}(x)$ is a topological basis for $X$. So we proved 4.

As you can observe unfortunately I can't prove the second statement and then in the first statement it result that $X$ is cozero-set and $\varnothing$ is zero-set whereas I proved the contrary. So I ask to prove the second point and to explain if what it is written in the final part of first point is a typo. Then I ask if the proof of 3 and 4 points is correct.
Could someone help me, please?
 A: 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.
A: 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.
