Does $ \mathfrak{A} \models (\exists x) \phi $ imply that $ \phi[x / t] $ for some term $ t $? If $ \mathfrak{A} \models (\exists x) \phi $, does it mean that $ \mathfrak{A} \models \phi[x / t] $ for some term $ t $?
I think that the answer is ‘yes’. After all, it is sufficient to let $ t = x $.
Am I wrong?

Generally, it is not true. It is true if $\mathfrak{A} $ contains such closed term $t$ ( function symbol) that $[t]^\mathfrak{A} =d  $ where $ (\mathfrak{A}, \delta) \models \exists x \phi, \delta(x) =d$
Yeah?

EDIT : for Berrick Caleb Fillmore
@Berrick Caleb Fillmore, you said that 

No matter ow you assign an element of || to each variable, you’re
  going to get all  or all . In general, the truth value of a
  sentence (i.e., its validity in the model $\mathfrak{A}$) is
  independent of your choice of valuation, because there aren’t any free
  variables that could possibly cause the truth value to fluctuate.

So, let's see how I reasoning about it
Let $\phi = \exists x p(x)$ . Let $\mathfrak{A} = (A, \Sigma^f, \Sigma^r) $ be a model for $\phi$. $A = \{1,2\}, \Sigma^f = \emptyset, \Sigma^r = \{p(1)\}$
Let $\delta_1 (x) = 1, \delta_2 (x) = 2$.
Now, we see that $(\mathfrak{A}, \delta_1) $ satisfies $\phi$ and $(\mathfrak{A}, \delta_2) $ satisfies $\phi$ doesn't. So it does mean that it depends on valutaion. 
Please help me understand where I am wrong- I know that you are right, of course. 
Thanks in advance :).
 A: We have to take care of the fine details of the semantical specifications ...
We can see :


*

*Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 84, for the definition of variable assignement function :



$s : \text {Var} \to |\mathfrak A|$,

used to define what it means for $\mathfrak A$ to satisfy $\varphi$ with $s$ :


$\mathfrak A \vDash \varphi[s]$.


Informally, we have that $\mathfrak A \vDash \varphi[s]$ iff the interpretation of $\varphi$ determined by $\mathfrak A$, when to the free occurrences of the variable $x$ the "object" $s(x)$ is "assigned" as denotation, is true.
In this case, the obvious inductive clause for $\exists$ will be :

$\mathfrak A \vDash \exists x \varphi [s]$ iff for some $d \in |\mathfrak A|$, we have $\mathfrak A \vDash \varphi [s(x|d)]$,

where $s(x|d)$ is the function which is exactly like $s$ except for the fact that at the variable $x$ it assignes the value $d$.
Intuitively, a free variable $x$ acts a a pronoun of natural language: to assert e.g. "$x$ is red" il like to assert "it is red". 
What is their meaning ? it depends on the context: if with "it" I'm pointing at my shirt, than the assertion is true. If instead I'm pointing at the book on the table, then the assertion is false.
The variable assignment function is a way to "formalize" the "context disambiguation" mechanism of natural language. Thus, the truth-value of "$x$ is red" depends on the denotation assigned to the variable $x$ by the function $s$.

We may have a different approach, where the semantical notions, like true, are defined only for sentences, i.e. closed formulae.
See : Joseph Shoenfield, Mathematical Logic (1967), page 19.
In this case, we consider closed terms (i.e. "names" for the objects of the domain) and the semantical clause for $\exists x \phi$ amounts to considering a "substitutional instance" $\phi[t/x]$ for some closed term $t$.
For details, see also : 
-George Boolos & John Burgess & Richard Jeffrey, Computability and Logic (4th ed - 2002), page 117 :

Let us say that in the interpretation $\mathcal M$ the individual $m$ satisfies $F(x)$, and write

$\mathcal M \vDash F [m]$,

to mean ‘if we considered the extended language $L \cup \{ c \}$ obtained by
  adding a newconstant $c$ in to our given language $L$, and if among all the extensions of our given interpretation $\mathcal M$ to an interpretation of this extended language we considered the one $\mathcal M^c_m$ that assigns to $c$ the denotation $m$, then $F(c)$ would be true’:

$\mathcal M \vDash F[m]$ if and only if $\mathcal M^c_m \vDash F(c)$.


