In which way are these logical statements similar to each other? 
*

*If x is even, then x is not divisible by 5.

*Every even integer is not divisible by 5.


Alright so the original problem is for me to determine a counterexample if these are false. I already found a counterexample for #1 which is x = 10. So can I use the same counterexample for question #2 since from my eyes they are referring to the same set of numbers? I'm trying to understand the answer of both to see how logically equivalent they are to each other (if they are).
 A: If we agree to formalize 1) as :

$Even(x) \to \lnot Div_5(x)$

and 2) as :

$\forall x (Even(x) \to \lnot Div_5(x))$

we have to take some care when speaking of logical equivalence.
According to some textbook, see e.g. Dirk van Dalen, Logic and Structure (5th ed - 2013), page 67, the definition of truth for a formula $\varphi$ in a structure $\mathfrak A$ is defined primarely for sentences (i.e. "closed" formulae) and then applied to open ones trough their "closure" :

$\mathfrak A \vDash \varphi$  iff $\mathfrak A \vDash Cl(\varphi)$.

Having that $\varphi$ is a semantic consequence of $\Gamma$ if $\Gamma \vDash \varphi$, i.e. when $\varphi$ holds in each model of $\Gamma$, 1) and 2) are entailing each other, and thus they are logically equivalent.
If instead we define, as in Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 88 :

Let $\Gamma$ be a set of wffs, $\varphi$ a wff. Then $\Gamma$ logically implies $\varphi$, written $\Gamma \vDash \varphi$, iff for every structure $\mathfrak A$ for the language and every function $s : Var \to |\mathfrak A|$ such that $\mathfrak A$ satisfies every member of $\Gamma$ with $s, \mathfrak A$ also satisfies $\varphi$ with $s$ (i.e. for every $\mathfrak A$ and $s$ : if $\mathfrak A \vDash \psi[s]$, for every $\psi \in \Gamma$, then $\mathfrak A \vDash \varphi[s]$ ),

we have that 1) $\nvDash$ 2). 
To show it, consider a variable assignment function $s$ such that $s(x)=11$; clearly $Even(11) \to \lnot Div_5(11)$ is true (because $F \to T$ is $T$), and thus : $\mathbb N \vDash (Even(x) \to \lnot Div_5(x)[s]$, while of course : $\mathbb N \nvDash \forall x (Even(x) \to \lnot Div_5(x))[s]$, i.e. :

$(Even(x) \to \lnot Div_5(x) \nvDash \forall x (Even(x) \to \lnot Div_5(x))$.

