You're correct for 1, 2, and 3. Below are some sketches for how to do the others.
To see why 2 is not finitely axiomatizable, you can take an ultraproduct of $\mathbb{Z}_p$ for $p\in{}\mathbb{N}$ prime. This is an infinite group, so the complement is not closed under ultraproducts, which means that the class of infinite groups is not finitely axiomatizable.
For 4, note that in a torsion group, each element has finite order. Let $C$ be the class of torsion groups and $T$ the theory. There are members of this class with elements of arbitrarily large order (look at $\mathbb{Z}_n$), so you can write sentences that say "there exists an element of order $n$" for each $n\geq{}1$. Then let $T'=T\cup{}\{{}\phi_n\}{}$ for each $n$. Since any finite subset of $T'$ is consistent, by compactness there is a model of $T'$ with an element of infinite order, thus is not a torsion group. So it is not axiomatizable.
For 5, we can do something similar to 2. For torsion-free groups we can include sentences $\phi_n$ that say the only element raised to the $n$th power that equals $e$ is $e$. However it is not finitely axiomatizable since we can take an ultraproduct of groups that have torsion elements and get a torsion free group. For example, take the ultraproduct of $\mathbb{Z}_p$. The structure with universe $\mathbb{Z}_p$ is a torsion-group for each $p$, but the ultraproduct will not be one.
To show a class is axiomatizable, you can find a set of axioms for it. For instance, $E$ being an equivalence relation is axiomatizable by the usual three axioms: $$ \forall x xEx\\\forall x,y(xEy\leftrightarrow yEz)\\\forall x,y,z(xEy\land yEz\to xEz).$$ Another property that is axiomatizable is $E$ being an equivalence relation with infinitely many equivalence classes. You can write down the property $\phi_3$ that there are three or more equivalence classes as $$ \exists x,y,z(\lnot x Ey\land \lnot x E z\land \lnot y Ez)$$ (this says $x,$ $y$, and $z$ are in different equivalence classes). You can similarly write down an axiom $\phi_4$ that says $E$ has greater than $4$ equivelence classes, and similarly a $\phi_n$ for any $n.$ Then the class of equivalence relations with infinitely many equivalence classes is axiomatized by the three equivalence axioms, plus the axioms $\{\phi_n:n\in\mathbb N\}.$
As your professor says, a good way to show something isn't axiomatizable is to use compactness. For instance, the class of equivalence relations with finitely many equivalence classes is not axiomatizable. To show this, we assume $\Sigma$ axiomatizes it and derive a contradiction. We add a countably infinite set of new constant symbols $\{c_1, c_2,\ldots\}$ to the language. Then we consider the set of axioms $\Sigma' = \{\lnot c_i E c_j: i\ne j\}$ that say the $c_i$ are all in different equivalence classes.
Let $T=\Sigma\cup \Sigma'.$ We can see that any finite subset of $T$ has a model, since any finite subset of $\Sigma'_0\subseteq \Sigma'$ can be satisfied by an equivalence relation with finitely many equivalence relations provided we make the number of classes large enough to assign all the $c_i$ that appear in $\Sigma_0'$ to different equivalence classes. Thus compactness says $T$ is satisfiable. Yet $T$ cannot be satisfiable since it is contradictory: $\Sigma$ says there are finitely many equivalence classes, but $\Sigma'$ implies there are infinitely many.
For another example that follows a similar pattern, we can show "$<$ is a well-ordering" is not axiomatizable. Assume for contradiction that $\Sigma$ axiomatizes it, and as before add $\{c_n:n\in\mathbb N\}$. Let $\Sigma'$ contain the sentences "the ordering has a greatest element and every element that isn't least has an immediate predecessor" (write these out!) as well as $c_i\ne c_j$ for all $i\ne j.$
Again, any finite subset of $T=\Sigma\cup \Sigma'$ is satisfiable. Simply find a finite ordered set that is large enough to assign the $c_i$ that appear in our finite number of sentences to different elements. Any finite set is well-ordered (so satisfies $\Sigma$) and has a greatest element and an immediate predecessor for every element except the least (so satisfies the rest of $\Sigma'$). So by compactness $\Sigma\cup \Sigma'$ has a model. But $\Sigma\cup\Sigma'$ says the model is an infinite well-ordered set that has a greatest element, and such that every element except the least has an immediate predecessor, and no such well-ordered set exists.
A good pattern we can abstract from this last example is that if we can find some axioms that have arbitrarily large finite models in the class but no infinite models in the class, then the class is not axiomatizable. (This also explains how we came up with the ordering properties in $\Sigma’,$ which probably seemed like they were pulled out of a hat at first glance.)
Best Answer
Hint for the first question: Let $\mathcal L$ be a language and let $T$ be any $\mathcal L$-theory with arbitrarily large finite models. Consider the language $\mathcal L'=\mathcal L\cup\{c_n\mid n\in\Bbb N\}$, where every $c_n$ is a constant symbol and the $\mathcal L'$ sentences $$\varphi_n=\bigwedge\limits_{\substack{1\leq i,j\leq n \\ i\neq j}}\neg(c_i=c_j)$$
so that intuitively $\varphi_n$ is saying "there are at least $n$ different elements. Is $T\cup\{\varphi_n\mid n\in\Bbb N\}$ finitely satisfiable? What does the compactness theorem tell you now?
Hint for the second question: Look again at the $\varphi_n$ from my first hint.
Hint for the third question: Suppose that $T$ is a finitely axiomatizable theory and that $\Gamma$ is an axiomatization of $T$. Can you show that there must be a $\Delta\subset\Gamma$, with $\Delta$ finite, such that $\Delta$ is an axiomatization of $\Gamma$? Now think again about your axiomatization from the second question.