The predicates are Students, Answers, and Questions.
So, you need two variables: say $x$ and $y$ for all but the 3rd translation:
Let $Sx$ denote "$x$ is a Student";
Let $Axy$ denote "$x$ answered $y$";
Let $Qy$ denote "$y$ is a question."
In each case you are going to need the existential quantifier to assert or deny the existence of a student and the universal quantifier to convey that we are making a statement about all questions.
$(1)$ None of the students answered all of the questions.
$(1)$ is the denial of existence. We can keep both quantifiers needed for this statement at the beginning by asserting the equivalent of $(1)$:
There does not exist an x such that for all y [if (x is a student and y is a question) then (Axy)].
Can you try working with translating that using symbolic logic?
- It is helpful, sometimes, to use "pseudo logic" as an intermediate step going from english to logic, or vice versa. Pay special attention to the parentheses.
$(2)$ At least one student answered all of the questions.
For $(2)$ we assert the existence of a student (i.e. "there exists at least one $x$ such that for all y, (if x is a student and y is a question, then....
$(3)$ One student answered all of the questions.
Here you need to assert that there exists one and only one student such that...
If you have studied the "existence-of-a-unique" quantifier $\exists !$, then your task is easy: you need only replace on quantifier used in $2$
If not, we can still express uniqueness using only the symbols $\exists, \;\;\forall,\;\; \text{and}\;\;=.$ We want to
Here you can still use your translation of $(2)$ - but you need to qualify it with "for all z (if z is a student and z answered all the questions, then z must be x)."
See if the following makes sense:
$(3)\;\;\exists x \forall y\{[(Sx \land Qy) \rightarrow Axy] \land \forall z [(Sz \land Azy) \rightarrow (z=x)]\}.$