The last (decimal) digit of a number is the number's residue modulo $10$, so the general kind of problem you need to solve is
Given a power tower $a_0^{a_1^{\vdots^{a_n}}}$ and a small modulus $m$, compute $$a_0^{a_1^{\vdots^{a_n}}} \bmod m$$
The first step is of course to rewrite this to
$$(a_0\bmod m)^{a_1^{\vdots^{a_n}}} \bmod m$$
using general facts about modular arithmetic. Now in the particular case that $a_0$ is coprime to $m$, Euler's theorem allows us to rewrite this to
$$ (a_0\bmod m)^{\bigl(a_1^{\vdots^{a_n}} \bmod \varphi(m)\bigr)} \bmod m$$
in which the exponent is now a simpler instance of the original problem -- simpler because the tower is one level shorter and $\varphi(m)$ is less than $m$ (unless $m=1$ in which case anything mod $m$ is $0$ anyway and you can throw the entire tower away).
Once you have reduced the exponent you can raise $a_0\bmod m$ to it by standard techniques such as exponentiation by squaring -- or just by winging it if $m$ is as small as 10 and you have 32-bit arithmetic.
If $a_0$ and $m$ are not coprime, then the slick theoretical approach is to factor $m$ and use the Chinese remainder theorem, but a kind of poor man's shortcut is this bastard offshot of Euler's theorem:
If $a$ and $m$ are arbitrary positive integers and $b\ge\varphi(m)$, then
$$ a^b \equiv a^{\varphi(m)+(b\bmod\varphi(m))} \pmod m$$
So if you can just recognize whether or not the upper tower is large (which is easy), you can reduce the task to raising to a number that is not larger than $2\varphi(m)$.
Since your starting $m$ is $10$, you don't need any fancy machinery for computing $\varphi(m)$ -- just hardcode a table for $m$ up to $10$. (In fact the only $m$ you will need are $10,4,2,1$).
This is not a complete answer to all of your questions. This is to show you some things you need to investigate. The first question is answered. The second question has an example. I do not know complete answers to the third and fourth questions, but I give a try on explaining your plot of $m=61$.
From your last sentences, it looks like you are interested in the case when $m$ is a prime. Let $m=p$ be an odd prime. Then consider $p\equiv 1$ mod $4$, and $p\equiv 3$ mod $4$.
In the former case $p\equiv 1$ mod $4$, we see the symmetric black dots. This is because the Legendre symbol at $-1$ is $1$. That is
$$
\left( \frac{-1}p \right)=1.
$$
This means $-1$ is a square of something in $\mathbb{Z}/p\mathbb{Z}$. Suppose $x\equiv y^2$ mod $p$, then we have $-x \equiv z^2$ mod $p$ for some $z\in\mathbb{Z}/p\mathbb{Z}$.
Your example $m=61$ is a prime that is $1$ mod $4$. Thus, we have a symmetric black dots.
In general when $p$ is an odd prime, the image of the square mapping is
$$\{ x^2 \ \mathrm{mod} \ p| 0\leq x \leq \frac{p-1}2 \}.$$
Note that the black dots represent image of the square mapping.
Thus, the number of black dots is $\frac{p+1}2$. In your example of $m=61$, we have $31$ black dots.
Now we use a primitive root $g$ in $\mathbb{Z}/p\mathbb{Z}$. Then any element $x\in \mathbb{Z}/p\mathbb{Z} - \{0\}$, we have some integer $a$ such that $x\equiv g^a$ mod $p$. Then a cycle formed by square mapping which includes $x$ can be written as
$$
\{g^{a\cdot 2^k} \ \mathrm{mod} \ p| k=0, 1, 2, \ldots \}.
$$
To see if we have cycles, try solving
$$
a\cdot 2^k \equiv a \ \mathrm{mod} \ p-1.
$$
In your plot of $m=61$, we have a primitive root $g=10$ and the following are cycles of length greater than $1$. All of these should be considered with modulo $61$.
$$
(g^{20} g^{40}),
$$
$$
(g^4 g^8 g^{16} g^{32}),
$$
$$
(g^{12} g^{24} g^{48} g^{36}),
$$
$$
(g^{56} g^{52} g^{44} g^{28})
$$
I am not sure if you consider these as cycles, because there can be numbers in front of these, such as
$$
g^5 \mapsto g^{10} \mapsto g^{20},
$$
and comes in to the cycle $(g^{20} g^{40})$.
Best Answer
Here is a realisation that helped me out back when I was learning modular arithmetic: Don't think of the symbol $\!\!\!\pmod k$ as an operation, but rather as part of the relation $\equiv$. For instance, $$ 7\equiv 2\pmod 5 $$ doesn't mean that if you take $7$ and apply the "modulo five" operation to it, you get $2$ (or the other way around). It rather means that $7$ and $2$ are in some sense "five-similar", i.e. they are both equally far away from any multiples of $5$. Think $7=_5 2$, if that makes any sense to you.
That way, you can also use, for instance $$ 1974 \equiv -1\pmod 5 $$ which might be helpful since $-1$ is often much easier to handle than $4$ in a lot of arithmetic, like if you were told to calculate $1974^{37}$ modulo $5$.
PS. The phrase "five-similar" is just something I came up with right now, and is not in any way conventional terminology. Unless you explain it, there is real risk that no one will understand it.