Here is a rough idea of the proof of 1. I'll highlight (with italics) concepts that you may not know now, but will be useful to learn in number theory, which hopefully will be motivated by this question.

Note that $a \in \operatorname{Hom}_K( T_\ell(E_1), T_\ell(E_2))$ induces $a_n : E_1 [\ell^n] \to E_2[\ell^n]$ for each $n$. We want to show, given such an $a \neq 0$, that there exist one or more homomorphism $E_1 \to E_2$ whose induced actions on the Tate module, in some $\mathbb Z_\ell$-linear combination, give $a$.

It's important to note that given just a single $a_n : E_1 [\ell^n] \to E_2[\ell^n]$, we cannot reach the desired conclusion. In fact, there can be such an $a_n$ for $n$ very large but no nonconstant maps $E_1 \to E_2$ at all.

So we have to take advantage of the fact that $a_n$ exists for infinitely many $n$.

We will do this by contrasting this infinitude against something else we know to be finite. For $K$ finite, this finiteness is going to ultimately boil down to the finiteness of $K$. For $K$ a number field, it's going to be subtler, but ultimately boil down to the fact that there are finitely many elements of bounded size (at each infinite place) in the ring of integers of $K$.

For each $n$, we can make an *abelian surface* that remembers the data of $a_n$, by writing $A_n= E_1 \times E_2 / \{ (x,y) \in E_1[\ell^n] \times E_2[\ell^n ] \mid y = a_n(x) \} $.

A priori, this appear to be infinitely many different abelian surfaces. The key claim to prove is that there are actually only finitely many different surfaces $A_n$ up to isomorphism.

Once we prove this, we will immediately get a bunch of isomorphisms $A_n \to A_m $ which combined with the natural maps $E_1 \to A_n$ and $A_m\to E_2$ will give us a lot of maps $E_1 \to E_2$. We have to do some algebra to show that we can combine these maps to recover $a$.

But the main arithmetic step is proving the finiteness of isomorphism classes.

For $K$ finite, we can first think about how we would solve the problem if the $A_i$ were elliptic curves. Then they would each have a $j$-invariant in the field $K$. There are infinitely many curves and only finitely many different $j$-invariants, and it's not so hard to see there can be finitely many (in fact, at most $6$) curves over a finite field with a given $j$-invariant, up to isomorphism.

In higher dimensions, the role of the $j$-invariant is portrayed by the coordinates of the *moduli space of abelian varieties*, which similarly parameterize isomorphism classes of abelian varieties. Since there are finitely many values for each coordinate, the moduli space has finitely many points. However, the moduli space only parameterizes abelian varieties with a *principal polarization*, so we use Zarhin's trick, which involves checking that $A_n^4$ has a principal polarization and using finiteness statements for higher-dimensional abelian varieties.

For $K$ a number field, it is much more complicated, even for elliptic curves. To get finitely many $j$-invariants, we have to show the numerator and denominator of the $j$-invariant are bounded. An equivalent statement is that the *height* of the $j$-invariant is bounded. To run the argument for abelian surfaces, we have to define a suitable *height function* on the moduli space, and show it is bounded on the sequence $A_n$.

The correct geometric method for defining a height function for this problem is *Arakelov theory*, which allows constructing height functions with much more precise properties than the elementary *Weil's height machine*. Faltings defined a specific height function that behaves very well under isogenies, like the isogenies between the $A_n$, in particular because they arise from a *$p$-divisible group*, enabling the use of *$p$-adic Hodge theory* to control the change in local terms.

verydifficult results, especially the latter. I don't know much of the history of 2, but I believe they are of comparable difficulty. $\endgroup$