The Noether-Deuring Theorem
My answer here will only answer question (1) above. The following is sometimes known as Zariski's Lemma:
Corollary 5.24 (Atiyah - Macdonald): Let $k$ be a field and $B$ a finitely generated $k$ algebra. If $B$ is a field then it is a finite (and hence algebraic) extension of $k$.
Proof: Our proof is different to that given in Atiyah - Macdonald. By Noether Normalization there exists a non-negative integer $n$ and algebraically independent elements $y_1,\ldots,y_n \in B$ such that $B$ is an integral extension of the polynomial algebra $k[y_1,\ldots,y_n]$. By proposition 5.7 of Atiyah - Macdonald we deduce that $k[y_1,\ldots,y_n]$ must be a field. However this can happen iff $n = 0$ which is to say $B$ is a finitely generated $k$ - algebra that is also integral (or algebraic) over $k$. Then
$$B = k(\alpha_1,\ldots,\alpha_n)$$
with each $\alpha_i$ of degree $k_i$ say. Consequently $\dim_k B \leq \prod k_i$; in particular $B$ is a finite extension of $k$.
to Question 3. Let me sketch a solution to the exercise which circumvents your troubles with infinite field extensions. It is not going to be a nice argument, but then again Noether-Deuring is a technical lemma to begin with, and IMHO should be avoided whenever possible.
Part 1 of the exercise easily follows from Part 2, because Part 2 shows that each of $V$ and $W$ is a direct summand of the other, and therefore $V\cong W$ (because no miracles happen in finite dimensions). So I'll only answer Part 2.
Case 1: The field extension $l/k$ is finite. Let $n=\left[l:k\right]$. Then, $V\otimes_k l\cong V^n$ and $W\otimes_k l\cong W^n$ as $A$-modules, as you proved. Thus, as $A$-modules, $W^n \cong W\otimes_k l \cong Y \oplus \left(V\otimes_k l\right) \cong Y\oplus V^n$.
Now, let us write each of $V$, $W$ and $Y$ as a direct sum of indecomposable $A$-modules. For each indecomposable $A$-module $I$, let $r_I\left(V\right)$ be the number of times $I$ appears in the decomposition of $V$, let $r_I\left(W\right)$ be the number of times $I$ appears in the decomposition of $W$, and let $r_I\left(Y\right)$ be the number of times $I$ appears in the decomposition of $Y$. These three numbers are well-defined because they don't depend on the choice of decompositions (since the Krull-Schmidt theorem says that in any decomposition of a finite-dimensional module into indecomposables, the multiset of the isomorphism classes of the indecomposables involved doesn't depend on the choice of the decomposition). If we can prove that $r_I\left(V\right) \leq r_I\left(W\right)$ for every indecomposable $I$, then we conclude that $V$ is a direct addend of $W$ as $A$-module (because the indecomposables in the decomposition of $V$ all appear in the decomposition of $W$ at least as often), so we are done (in Case 1, at least).
To prove that $r_I\left(V\right) \leq r_I\left(W\right)$ for every $I$, we notice that every $I$ satisfies
$n\cdot r_I\left(W\right) = r_I\left(W^n\right) = r_I \left(Y \oplus V^n\right)$ (since $W^n\cong Y\oplus V^n$) $= r_I\left(Y\right) + r_I\left(V^n\right) \geq r_I\left(V^n\right) = nr_I\left(V\right)$
and thus $ r_I\left(W\right) \geq r_I\left(V\right)$. This solves Part 2 of the exercise in Case 2.
Case 2: The general case.
We are looking for two $A$-module homomorphisms $i:V\to W$ and $p:W\to V$ satisfying $p\circ i=\mathrm{id}$. (In fact, if we can find such $i$ and $p$, then it becomes clear that $V$ is a direct addend of $W$ as $A$-module.) Since $V$ and $W$ are finite-dimensional, we can identify the $k$-vector spaces $V$ and $W$ with $k^n$ and $k^m$ for some nonnegative integers $n$ and $m$, and then $k$-linear maps $V\to W$ can be identified with $m\times n$-matrices over $k$, while $k$-linear maps $W\to V$ can be identified with $n\times m$-matrices over $k$. The conditions that the two maps $i$ and $p$ be $A$-module homomorphisms translate into a system of linear constraints on the entries of these two matrices; and of course, we can WLOG assume that there are only finitely many constraints (because there are only finitely many entries, so if there are infinitely many constraints we can throw away almost all of them for redundancy). The equation $p\circ i=\mathrm{id}$ translates into a system of polynomial equations for the entries of these two matrices. So we are looking for $mn+nm=2nm$ elements of $k$ which satisfy certain linear and certain polynomial equations. Since linear equations are also polynomial, what we are looking for is thus simply a solution for a fixed system of finitely many polynomial equations over $k$ in finitely many indeterminates.
However, because we know that $V\otimes_k l$ is a direct addend of the $A\otimes_k l$-module $W\otimes_k l$, we know that such a solution can be found over $l$ (because tensoring with $l$ means extending scalars to $l$).
Now, a general fact from algebraic geometry says that
(*) if $l/k$ is a field extension, and some given system of finitely many polynomial equations over $k$ in finitely many indeterminates has a solution over $l$, then there exists a finite field extension $m$ of $k$ such that this system has a solution over $m$.
Applying this to our system, we get a finite field extension $m$ of $k$ such that our system has a solution over $m$. This solution translates into a witness that $V\otimes_k m$ is a direct summand of $W\otimes_k m$ as an $A\otimes_k m$-module. But since $m/k$ is finite, we can now apply Case 1 to $m$ in lieu of $l$, and conclude that $V$ is a direct summand of $W$ as an $A$-module. This solves Part 2 in the general case.
to Question 2. I am very interested in this myself; it is similar to MathOverflow question #48909 which is still unanswered.