Regarding number bases less than $1$ (Question 1), suppose you have a base-one-tenth number system. The first few integers in this system would be
$$1,2,3,4,5,6,7,8,9,0.1,1.1,2.1,3.1,4.1,5.1,6.1,7.1,8.1,9.1,0.2$$
The reason $0.1$ is the next integer after $9$ is that the base
of this number system, $b$, has the value we would normally call
$\frac{1}{10}$ (writing in base ten),
and the first place after the decimal point has place value $b^{-1}$.
And of course $\left(\frac 1n\right)^{-1} = n$,
for any $n$.
This is kind of a silly example since you can just write all your numbers
backwards and put the decimal point between the ones' place and the
tens' place. So it is not too surprising that I've never seen this
system explained before.
Moving on to Question 1.5,
a much more interesting system is the base-$\phi$ number system.
The number $\phi$ is also known as the Golden Ratio:
$\phi = \frac12(1 + \sqrt 5)$.
It turns out that all the powers of $\phi$ have the form
$\frac12(a + b\sqrt 5)$ for some integers $a$ and $b$,
and lots of nice cancellation can occur.
So if we allow only the digits $0$ and $1$
in this number system (these are the only non-negative integers
less than $\phi$), we can write the first few integers as
$$1, 10.01, 100.01, 101.01, 1000.1001, 1010.0001$$
But regarding Questions 3 and 4:
whatever system it is written in, a whole number is a whole number.
Since we're so used to using only whole-number-based systems,
the numbers written in base-one-tenth and base-$\phi$ in the lists above
(the "first few integers" in each base)
may not look like whole numbers, but nevertheless
that's what every single one of them is.
Or to put it another way,
each whole number is produced by adding $1$ to the number before it.
That is (essentially) the way the whole numbers are defined.
The numbering system you use may have some bizarre effects on how
you "carry" digits when adding $1$ to the previous whole number
causes the ones' place to "roll over", but as long as you do the
arithmetic operation correctly you will get the same actual number
as you would have gotten in any other base.
Likewise any other mathematical facts involving integers or fractions
are exactly the same in an number base; they may just look different
due to the different numerical notation.
Assuming Polignac's conjecture we will always be able to find two primes $(c,d)$ such that
$$
\left\lfloor{ \frac{a}{10}} \right\rfloor - \left\lfloor{\frac{b}{10}}\right\rfloor = -\left(\left\lfloor{\frac{c}{10}}\right\rfloor - \left\lfloor{\frac{d}{10}}\right\rfloor\right)
$$
(the distance between $a$ and $b$ along the $x$-axis is equal to the negative of the distance between $c$ and $d$) and
$$
a = b,\; \; c=d \; \mod 10
$$
($a$ and $b$, and $c$ and $d$, end in the same digits).
This defines an isosceles trapezium, which is always a cyclic quadrilateral (a quadrilateral such that a circle can be drawn with its 4 vertices.
If $a = b \mod 10 \;$, the above argument still probably holds, but I have not found a proof.
Best Answer
Hint $ $ The number of residues coprime to $n> 2$ is even: $ $ negation reflection $\,x\mapsto -x\pmod {\!n}\,$ partitions them into pairs (since it has no fixed points: $\,-a\equiv a\,\Rightarrow\, n\mid 2a,\,$ contra $(n,a)=1)$.
Remark $\ $ Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.