I don't think you can say that because there's no infinite structure, there is a maximal structure. In fact, that is pretty much what you're trying to prove. It's certainly conceivable at the outset that there could be a structure of every finite size, no matter how big, but no infinite structure.
The key to both problems is a compactness argument where you add a set of new constant symbols $\{c_i:i\in I\},$ as well as the axioms $c_i\ne c_j$ for all $i,j\in I.$ Then you show that every finite subtheory of this extended theory has a model by finding a model with more elements than $c_i$ that appear in the subtheory and assigning distinct elements to be interpretations of those $c_i.$ Then by compactness, the theory has a model, which by construction must have cardinality $\ge |I|.$
The notation $Mod(\Sigma)$ is not something I've seen very much (though even if you hadn't defined it, it would have been pretty clear from context what it was.) The most effecient expression I can think of is "the class of all models of $\Sigma$".
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I will explain the compactness argument in more detail, for the particular case of problem 2 (as I've said, 1 is very similar). Let $\Sigma$ be a $L$-theory that has models of every finite size. Now, let $\{c_i: i\in \mathbb N\}$ be a countably infinite collection of distinct constant symbols that do not appear in $L.$ Consider the extended language $L' = L\cup \{c_i: i\in \mathbb N\}$ and the $L'$-theory $$ \Sigma'=\Sigma \cup \{c_i\ne c_j: \mbox{$i,j\in \mathbb N$ and $i\ne j$}\}.$$
We claim $\Sigma'$ has a model. By compactness, it suffices to show that any finite subtheory of $\Sigma'$ is satisfiable. So let $\Sigma_0$ be such a finite subtheory. Then only a finite number $$ \{c_{i_k}:k\in \{1,2,\ldots, n\}\}$$ of the new constant symbols occur in $\Sigma_0.$ Then let $M$ be a model of $\Sigma$ with $n$ or more elements (which exists by hypothesis). Then $M$ becomes a model of $\Sigma_0$ when we interpret each of the $c_{i_k}$ as distinct elements of $M,$ since distinctness means it will satisfy any statement of the form $c_{i_k} \ne c_{i_{k'}}$ that might appear in $\Sigma_0.$
So let $N$ be a model of $\Sigma'.$ Clearly $N$ is a model of $\Sigma.$ We can also see that $N$ is infinite, since if it were finite, then one of the $c_i\ne c_j$ axioms in $\Sigma'$ would have to fail by pigeonhole.
For the first question, the point is that there are basic algebraic/combinatorial phenomena which can only occur in infinite structures. The simplest example is an injective non-surjective function: in the language with a unary function symbol $f$, the sentence $$[\forall x, y(x\not=y\implies f(x)\not=f(y))]\wedge[\exists x\forall y(f(y)\not=x)]$$ does have models but does not have any finite models. Another easy example is linear orders without endpoints: in the language with a binary relation symbol $R$, the sentence asserting that $R$ is a linear order with no greatest (or least, if you prefer) element has only infinite models.
(There are also surprisingly deep examples, e.g. Wedderburn's little theorem says that the sentence saying "the universe is a non-commutative division ring" has only infinite models.)
For the second question, this is just a direct application of the theorem. Let $\mathbb{K}$ be the class of all finite structures (in some fixed language). If $\mathbb{K}$ were $EC_\Delta$, then there would be some set of sentences $\Gamma$ such that $\mathbb{K}$ is exactly the class of all models of $\Gamma$. But:
Do you see why such a $\Gamma$ must have arbitrarily large finite models? (HINT: how large can elements of $\mathbb{K}$ be?)
Do you see why such a $\Gamma$ cannot have infinite models? (HINT: how large can elements of $\mathbb{K}$ not be?)
This gives a contradiction, so our assumption that $\mathbb{K}$ was $EC_\Delta$ must have been wrong.
Best Answer
No, that doesn't work. It's hard to pin down exactly where it breaks since you haven't really given details about how you're applying compactness here, but in fact the class of infinite structures is closed under elementary equivalence. This is because for each finite $n$ there is a sentence $\varphi_n$ which is true in exactly the structures of size $\ge n$; if $\mathcal{A}$ is infinite and $\mathcal{B}$ is finite, then we have $$\mathcal{A}\models\varphi_{\vert\mathcal{B}\vert+1}\quad\mbox{but}\quad\mathcal{B}\not\models\varphi_{\vert\mathcal{B}\vert+1}.$$
Namely, $\varphi_n$ is the sentence $$\exists x_1,...,x_n(\bigwedge_{1\le i<j\le n}x_i\not=x_j).$$
Instead, we need to use the (downwards) Lowenheim-Skolem theorem. You've correctly observed that the class $C_\kappa$ of structures of cardinality $\ge\kappa$ is always closed under ultraproducts, and so we're looking for something which can "shrink" structures while preserving elementary equivalence. This (with $\kappa=\aleph_0$) is what you've tried to use compactness to do, but that didn't work. Instead, use downwards Lowenheim-Skolem (with $\kappa=\aleph_1$ - but $\aleph_{\omega^2+17}$, or indeed any uncountable $\kappa$, would also work): the class $\mathbb{U}$ of uncountable structures is trivially closed under ultraproducts, but by downwards Lowenheim-Skolem for each $M\in \mathbb{U}$ there is a countable (hence $\not\in \mathbb{U}$) structure $N$ with $M\equiv N$, so $\mathbb{U}$ is not closed under elementary equivalence.