Nested Quantification of exactly one. Suppose my domain is "All students in the class" and P(x, y):= x has emailed y.
So, how do i define:


*

*Every student has emailed exactly one student.

*Exactly one student has emailed every one.


A proper explanation would be really helpful. 
 A: "Exactly one" means "at least one, and at most one". 
We express "there is at least one $z$ such that $\phi(z)$" with 
$$
\exists z\, \phi(z).
$$
We express "there is exactly one $z$ such that $\phi(z)$" as follows:
$$
\exists z\, (\phi(z) \land \forall u\,(\phi(u)\to u=z)).\tag{*}
$$
This says, there is something $z$ such that $\phi(z)$ (at least one), and anything else $u$ such that $\phi(u)$ must equal $z$ (at most one).
Now for your two statements.
Every student has emailed exactly one student.
Using variables, the statement is: Every student $x$ has emailed exactly one student $y$. I take this to mean: For every student $x$, there is exactly one $y$ such that $x$ has emailed $y$. Using the analysis of "exactly one" above, we can render this symbolically as follows, using $\phi(y) := P(x,y)$ and universally quantifying:
$$
\forall x \exists y\, (P(x,y) \land \forall u\,(P(x,u)\to u=y)).
$$
Exactly one student has emailed everyone.
We want a formula for "$x$ has emailed everyone" (i.e. every student, presumably including self). $\phi(x) := \forall y\,P(x,y)$ says that. Now we only have to say there's exactly one such $x$:
$$
\exists x\, (\forall y\,P(x,y) \land \forall u\,(\forall y\,P(u,y)\to u=x)).
$$
