$c$ is a propositional constant in $K_{\mathcal{L}}$. Prove or falsify the following conclusions:
(1) $\models\exists x\alpha\rightarrow\alpha(x/c)$
(2)If $\models\exists x\alpha$, then$\models\alpha(x/c)$
I've seen answers in Does $ \mathfrak{A} \models (\exists x) \phi $ imply that $ \phi[x / t] $ for some term $ t $? and Why is it not true that if $\Phi \models \exists x \phi$ then there is a term $t$ with $\Phi \models \phi \frac{t}{x}$?, but they all give the counterexamples with special $\Phi$ or $\mathfrak{A}$. Without them, are the conclusions above still wrong? If so, how to find an appropriate counterexample?
In situations like these it helps to write out the explicit meta-level quantification in the thing you're trying to prove and to remember that $\models \varphi$ holds if and only if $M \models \varphi$ for all structures $M$.
For example (1) is really saying: for all well-formed formulas $\alpha(x)$, for all structures $M$, it holds that $M \models \exists x \mathop. \alpha(x)$. From there you instantiate your otherwise-featureless structure $M$ and well-formed formula $\alpha$ and do case analysis or set up an induction or argue in some way.
Also, note that cases with $\Phi \models \gamma$ are actually more general. $\Phi \models \gamma$ is equivalent to, for all structures $M$, if $M \models \Phi$, then $M \models \gamma$. Proving something about $\models \gamma$ is proving a special case when $\Phi$ is $\varnothing$.
(1) is false.
We want to disprove $\models (\exists x \mathop. \alpha(x)) \to \alpha(c)$.
Consider $x \neq c$ as $\alpha(x)$.
In a structure consisting of $\{c, d\}$ where $c$ interprets as itself, $\exists x \mathop. x \neq c$ holds and $c \neq c$ fails. Therefore, it is not the case that $\models (\exists x \mathop. \alpha(x)) \to \alpha(c)$.
(2) is false.
Let $\exists x \mathop. \alpha(x)$ be the well-formed formula $\exists x \mathop. (\forall z y. z = y) \lor x \neq c$.
In any given structure $M$, $\exists x \mathop. (\forall z y. z = y) \lor x \neq c$ is true. The domain can't be empty because we have a constant symbol. Additionally, if the domain has only one element, then $x$ can be $c$. However, if the domain has more than one element, $x$ must not be $c$. Assigning $x$ to some non-$c$ element is always possible when the domain has more than one element.
However, the sentence $\alpha(c)$ which is $(\forall z y. z = y) \lor c \neq c$ is true precisely when the structure has a domain with exactly one element in it, thus it is not a tautology.