Let's denote the tensor product of Hilbert spaces by $\widehat{\otimes}$, the tensor product of vector spaces by $\otimes$.
We have a natural bilinear map $K \times \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$ which sends $(a,(h_i))$ to $(a \otimes h_i)$. It extends to a linear map $K \otimes \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. This map is isometric because
$$\langle (a \otimes h_i),(a' \otimes h'_i) \rangle = \sum_i \langle a,a' \rangle \cdot \langle h_i,h'_i \rangle = \langle a,a' \rangle \langle (h_i),(h'_i) \rangle.$$
Hence it extends to an isometric linear map on the completion $K \widehat{\otimes} \bigoplus_i H_i \to \bigoplus_i (K \widehat{\otimes} H_i)$. One checks that its image is dense. The image is also closed since it is complete. Thus it is an isometric isomorphism.
In this answer, I will use $x_n$ as a sequence in $l^2$ and write $x_n(k)$ as the $k$-th member of that sequence.
The norm in the Hilbert space is given by $\|x\| = \sqrt{\langle x, x \rangle}$. We wish to show that if a sequence $\{ x_n \} \subset l^2$ is Cauchy, then it converges in $l^2$.
Suppose that $\{x_n\}$ is such a Cauchy sequence. Let $\{ e_k \}$ be the collection of sequences for which $e_k(i) = 1$ if $i=k$ and zero if $i\neq k$.
Then $\langle x_n, e_k \rangle = x_n(k)$. Notice that $$|x_n(k) - x_m(k)| = |\langle x_n - x_m, e_k \rangle| \le \|x_n-x_m\| \| e_k\| = \|x_n-x_m\|$$ for all $k$ (also note that this convergence is uniform over $k$). Therefore the sequence of real numbers given by $\{x_n(k)\}_{n\in \mathbb{N}}$ is Cauchy for each $k$, and thus converges. Call the limit of this sequence $\tilde x(k)$.
Let $\tilde x = (\tilde x(k))_{k\in\mathbb{N}}$. We wish to show that $\tilde x \in l^2$.
Consider $$\sum_{k=1}^\infty |\tilde x(k)|^2=\sum_{k=1}^\infty |\lim_{n\to\infty} x_n(k)|^2=\lim_{n\to\infty} \sum_{k=1}^\infty |x_n(k)|^2=\lim_{n\to\infty}\|x_n\|^2.$$
The exchange of limits is justified, since the convergence of $\lim_{n\to\infty} x_n(k)$ is uniform over $k$. Finally, since $\{ x_n \}$ is Cauchy, the inequality, $$| \|x_m\| - \|x_n\| | < \| x_m - x_n\|$$ implies that $\|x_n\|$ is a Cauchy sequence of real numbers, and so $\|x_n\|$ converges. Thus $\tilde x$ is in $l^2$.
Edit: Completing the proof as per the comments.
We have thus shown that $\tilde x$ is in $l^2$. $\tilde x$ is the most likely candidate for the Cauchy sequence to converge to, and it has been demonstrated to be in our space. What remains is to show that $$\| x_n - \tilde x\| \to 0$$ as $n \to \infty$.
We will utilize a generalized form of the dominated convergence theorem for series. This states that if $a_{n,k} \to b_k$ for all $k$, $a_{n,k} < d_{n,k}$ and $\sum_{k} d_{n,k} \to \sum_{k} D_k < \infty$, then $\lim_{n \to \infty} \sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty b_k$. (here $a_{n,k}, b_k, d_{n,k}, D_{k}$ are all real numbers)
Writing $$\| x_n - \tilde x\|^2 = \sum_{k=0}^\infty |x_n(k) - \tilde x(k)|^2.$$
We see that in this case $a_{n,k} = |x_{n}(k) - \tilde x(k)|^2$, $b_k = 0$, and we must find a $d_{n,k}$ that "dominates" $a_{n,k}$ to finish the proof.
Now note that $|x_n(k) - \tilde x(k)|^2 \le 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2$ and $$\lim_{n \to \infty} \sum_{n=0}^\infty ( 2 |x_n(k)|^2 + 2 |\tilde x(k)|^2) = \sum_{k=0}^\infty (2 |\tilde x(k)|^2 + 2 | \tilde x(k)|^2).$$ Recall that we demonstrated $\lim_{n \to \infty} \sum_{n=0}^\infty |x_n(k)|^2 = \sum_{n=0}^\infty |\tilde x(k)|^2$ in the first half. Thus $D_k$ is played by $4|\tilde x(k)|^2$ in this case.
Thus by the dominated convergence theorem we may conclude that $$\sum_{k=0}^\infty |x_n(k)-\tilde x(k)|^2 \to 0.$$
Best Answer
1) If $n\geq1$, it is clear that the closure of $\ell^2(A)$ contains the subspace of $\ell^2(H)$ of those sequences which vanish from the $n$th component onwards. It then contains the closure of the union of these subspaces, which is $\ell^2(H)$.
2) Nope, in general. If $(a_n)_{n\geq1}$ is a linearly independent sequence of elements of $A$ such which is in $\ell^(A)$, then it is not in the subspace $\ell^2\otimes A\subseteq\ell^2(A)$.
3 and 4) Yes. Proceed as in (1) to show this.