Another approach to the 6th part.
Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.
Clearly
$$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$
Let us try to show the opposite inequality.
For arbitrary function $f\in\operatorname{Bij}(\mathbb N,\mathbb N)$ we define $$\operatorname{Fix}(f)=\{n\in\mathbb N; f(n)=n\}.$$ (That is, $\operatorname{Fix}(f)$ is the set of all fixed points of the map $f$.)
Let us try to answer the question, whether for any
$A\subseteq\mathbb N$ it is possible to find a bijection ${f_A}:{\mathbb N}\to {\mathbb N}$ such $\operatorname{Fix}(f_A)=A$. Let us consider two cases.
If the complement of the set $A$ is infinite, i.e. $\mathbb N\setminus A=\{b_n; n\in\mathbb N\}$ (and we can assume that the elements $\mathbb N\setminus A$ are written in the increasing order, i.e. $b_1<b_2<\dots<b_n<b_{n+1}<\dots$ ), then we can define a function $f_A$ as follows:
$$f_A(x)=
\begin{cases}
x, & \text{if }x\in A, \\\
b_{2k+1}, &\text{if $x=b_{2k}$ for some $k\in\mathbb N$},\\\
b_{2k}, &\text{if $x=b_{2k+1}$ for some $k\in\mathbb N$}.
\end{cases}
$$
In the other words, we did not move the elements of $A$ and the elements from the complement were paired and we interchanged it pair.
Such map is a bijection from $\mathbb N$ to $\mathbb N$ such that $\operatorname{Fix}(f_A)=A$.
Next we consider the case that $\mathbb N\setminus A$ is finite.
If $\mathbb N\setminus A=\emptyset$, then $f=id_{\mathbb N}$ is a function fulfilling $\operatorname{Fix}(f_A)=\mathbb N=A$.
If the set $\mathbb N\setminus A$ is a singleton $\{a\}$, then is is impossible to find a bijection $f$ such that $\operatorname{Fix}(f)=A=\mathbb N\setminus\{a\}$. No element of $A$ can be mapped on $a$ (since all such elements are fixed). But $a$ cannot be mapped on $a$ since this would mean that $a$ is a fixed point.
But if the set $\mathbb N\setminus A$ has at least two elements, then it is possible to construct such a map. We again assume that the elements $b_k$ are written in the increasing order, i.e. $\mathbb N\setminus A=\{b_0,\dots, b_n\}$ and $b_0<b_1<\dots<b_n$.
We put $f_A(x)=x$ for $x\in A$, $f_A(b_k)=b_{k+1}$ for $0\le k<n$ and $f_A(b_n)=b_0$. (I.e. we made a cycle consisting of elements of $\mathbb N\setminus A$.)
This defines a bijection ${f_A}:{\mathbb N}\to{\mathbb N}$ such that $\operatorname{Fix}(f_A)=A$.
The assignment $A\mapsto f_A$ that we described above is a map from the set $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ consisting of all subsets of $\mathbb N$, whose complement is not a singleton, to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This map is injective; since from $f_A=f_B$ we get $A=\operatorname{Fix}(f_A)=\operatorname{Fix}(f_B)=B$.
From the set $\mathcal P({\mathbb N})$ with the cardinality $\mathfrak c$ we have omitted a countable set $\{\mathbb N\setminus\{a\}\in\mathbb N\}$. Therefore the cardinality of this difference $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ is again $\mathfrak c$.
So we found an injection from a set of cardinality $\mathfrak c$ to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This yields the opposite inequality
$$\mathfrak c\le|{\operatorname{Bij}(\mathbb N,\mathbb N)}|$$
and by Cantor-Bernstein theorem we get $|{\operatorname{Bij}(\mathbb N,\mathbb N)}|=\mathfrak c$.
Unfortunately, these countably infinitely many functions still only hit a small part of $2^{\Bbb N}$. We can do a very similar diagonalisation argument to show that no matter which functions $f_i$ you choose, there will always be some $s\in 2^{\Bbb N}$ that none of them hit.
This will require a bit more book-keeping than the standard diagonalisation argument for a single function. So it can look a bit messy. But if you keep in the back of your mind that the basic idea is basically the same, you should be able to transfer your understanding of the diagonal proof to this one.
Let $2^{\Bbb N}$ be the set of all countably infinite binary sequences, and assume that for each $i\in \Bbb N$, we have a function $f_i:\Bbb N\to 2^{\Bbb N}$ (we can require that these all be distinct, or even that all their ranges are disjoint, but there is no need for such requirements).
For simplicity, we establish the following notation: Given a binary sequence $t$, let $t_i$ be the $i$th entry.
Now for the proof. We set
$$
s_1 = 1-f_1(1)_1\\
s_2 = 1-f_1(2)_2\\
s_3 = 1-f_2(1)_3\\
s_4 = 1-f_1(3)_4\\
s_5 = 1-f_2(2)_5\\
s_6 = 1-f_3(1)_6\\
s_7 = 1-f_1(4)_7\\
s_8 = 1-f_2(3)_8\\
s_9 = 1-f_3(2)_9\\
s_{10} = 1-f_4(1)_{10}\\
\vdots
$$
The idea goes as follows: For the $i$th component of $s$, we take the $i$th component of some $f_m(n)$ and flip it. This ensures $s\neq f_m(n)$. Then we go through all possible $f_m(n)$ one by one in a manner that ensures that we eventually go through them all. In this case, that manner is to first do $f_1(1)$. Then $f_1(2)$ and $f_2(1)$. Then $f_1(3)$, $f_2(2)$ and $f_3(1)$. And so on.
Best Answer
Choose an enumeration of the finite subsets of $\Bbb N$. There are countably many such subsets so such an enumeration is possible.
For each finite subset of $S \subseteq \Bbb N$, there are only countably many maps $g: S \to \Bbb N \setminus \{ 1 \}$. There is a one-to-one correspondence between elements of $\mathscr A$ and ordered pairs $(S, g)$. There are only countably many possible choices for $S$ and for each $S$ only countably many choices for $g$. Therefore, there are only countably many such ordered pairs, so $\mathscr A$ is countable.
As noted in Mark Saving's answer, the problem with a diagonal argument is that you haven't proved (and can't prove) that the new function created using the diagonal technique is in fact also in $\mathscr A$ because you haven't proved that it takes on the value $1$ at all but finitely many arguments.