Translate the statement into English and something about the order. My question is the exact order for this statement below. I'm definitely clueless. Even I have no idea where I should start from.
Actually, what I wrote was "Every student has a computer or there is a friend who has a computer and is a friend with every student."
I did this way $∀x(C(x))∨ (∀x∃y(C(y)∧(F(x,y)))$
Is it fine?
Define $C(x)$ : " $x$ has a computer" and $F(x,y)$: "$x$ and $y$ are friends" where $x$ and $y$ are over the domain of students.
$∀x(C(x)∨∃y(C(y)∧(F(x,y)))$ 
 A: Let's translate it literally first:
$\forall x \,(C(x) ∨ \exists y \,(C(y)∧(F(x,y)))$
For all students x, $(C(x) ∨ \exists y \,(C(y)∧(F(x,y)))$
For all students x, student x has a computer $∨ \, \exists y \,(C(y)∧(F(x,y)))$
For all students x, student x has a computer or there exists a student y such that $(C(y)∧(F(x,y)))$
For all students x, student x has a computer or there exists a student y such that student y has a computer $∧ \, (F(x,y))$
For all students x, student x has a computer or there exists a student y such that student y has a computer and students x and y are friends. 

Or, more compactly: Every student has a computer or is a friends with a student who has a computer.
A: So it turns out my answer was wrong. Instead of an answer this will now become an explanation as to why.

Here was my original answer:

A faithful translation of your statement written in English is indeed
  what you wrote in using logical symbols bellow it. 
However; if you wished to produce a more compact statement that yields
  the same results; you could go with
∀x[∃y(C(y)∧(F(x,y))]
The only difference here is that the second statement does not double
  up on the case in which every student has a computer. The reason this
  occurs is because the $\exists$ operator permits the case in which
  every value is true.


And here is my explanation of what goes wrong.
Consider a world in which only one member exists: p.
In this world C(p) is true, and F(p,p) is false.
When we consider the first statement:

∀x(C(x))∨∀x(∃y(C(y)∧(F(x,y))))

Since there only exists one thing in W, this means that:
∀x(C(x)) in W
means C(p), which is true since C(p) is true.
Thus the whole first statement is true in W.
If we instead consider the second statement:

∀x(C(x)∨∃y(C(y)∧(F(x,y))))

we find that in W it means:
C(p)∨∃y(C(y)∧(F(p,y)))
which is true because C(p) is true.
Now if we were to consider my answer:

∀x[∃y(C(y)∧(F(x,y))]

we instead find that it means:
∃y(C(y)∧(F(p,y))
which is just C(p)∧F(p,p).
But F(p,p) is false. Thus the statement as a whole is false.
Therefore because there exists a world in which my statement is false and both statements in the OP are true; my statement is not equivalent.
A: Almost.



*

*Every student has a computer or there is a friend who has a computer and is a friend with every student.

*$\big(\forall x{\in}\mathrm S~C(x)\big)$ or there is a friend who has a computer and is a friend with every student.

*$\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big($ "who has a computer and is a friend with every student.$\big)\Big)$

*$\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big(C(y)$ "and is a friend with every student.$\big)\Big)$

*$\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big(C(y) \land (\forall x{\in}\mathrm S~F(y,x))\big)\Big)$

*$\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\forall x{\in}\mathrm S~\big( C(y)\land F(y,x)\big)\Big)$
Key point: the order of the existential and universal quantifiers cannot be swapped.

The statement $∀x(C(x))∨(∀x∃y(C(y)∧(F(x,y)))$, if we assume the domain of Students, translates as
"Either every student has a computer, or every student is friends with a student who has a computer."
A: Just for completeness, I shall prove that the two statements you have in your question are not equivalent (I also assumed that both are missing a bracket at the end):

(1) $∀x(C(x))∨(∀x∃y(C(y)∧(F(x,y))))$.
(2) $∀x(C(x)∨∃y(C(y)∧(F(x,y))))$.

Let $W$ be a world with exactly $2$ objects $p,q$ such that the following hold:

*

*$C(p)$, $\neg C(q)$.


*$\neg F(p,p)$, $F(q,p)$.
Then (1) is false in $W$ because both "$∀x(C(x))$" and "$∀x∃y(C(y)∧(F(x,y)))$" are false in $W$. To see that that second half is false in $W$, note that it in particular asserts that "$( \exists y\ ( C(y) \land F(p,y) )$" is true, but that is false because neither $p$ nor $q$ is a valid witness ($p$ fails since "$F(p,p)$" is false, and $q$ fails since "$C(q)$" is false).
But (2) is true in $W$ because it is true when applied to both $p$ and $q$ (for $p$ since $C(p)$, and for $q$ since $C(p)$ and $F(q,p)$).
Therefore (1) and (2) are not equivalent because they have different truth values in some world.
