For primes $p > 5$, the Binet formula applies mod $p$, in the following sense.
If $p \equiv \pm 1 \mod 5$, $5$ is a quadratic residue mod $p$, and $z^2 - z - 1$ has two roots $\phi_\pm = (1 \pm \sqrt{5})/2$ in ${\mathbb Z}_p$.
Then $F_n \equiv ((\phi_+^n - \phi_-^n)/\sqrt{5} \mod p$. In particular,
$F_{p-1}$ is divisible by $p$, so $\gcd(p, F_{p-1}) = p$.
If $p \equiv \pm 2 \mod 5$, $z^2 - z - 1$ doesn't have roots in ${\mathbb Z}_p$,
but it has roots in an extension field $GF(p^2)$. In this case it turns out that $F_{p+1}$ is divisible by $p$, so $\gcd(p, F_{p+1}) = p$.
EDIT: Note that we have $\phi_+ + \phi_- = 1$ and $\phi_+ \phi_- = -1$.
Since we're in characteristic $p$,
$\phi_+^p + \phi_-^p = 1^p = 1$, and $\phi_+^p\phi_-^p = (-1)^p = -1$ as well.
So $\phi_+^p$ and $\phi_-^p$ are also roots of $z^2 - z - 1$. But there are only two such roots. In the second case, we can't have $\phi_+^p = \phi_+$, because only elements of ${\mathbb Z}_p$ have the property $z^p = z$, so we must have $\phi_+^p = \phi_-$, and similarly $\phi_-^p = \phi_+$.
Then $$F_{p+1} = \dfrac{\phi_+^{p+1} - \phi_-^{p+1}}{\sqrt{5}} = \dfrac{\phi_- \phi_+ - \phi_+ \phi_-}{\sqrt{5}} = 0 \ \text{(as a member of $GF(p^2)$)}$$
i.e. $F_{p+1} \equiv 0 \mod p$.
The Fibonacci sequence represents a sort of "worse case" for the Euclidean algorithm. This occurs because, at each step, the algorithm can subtract $F_n\,\, $ only once from $F_{n+1} \,\,$. The result is that the number of steps needed to complete the algorithm is maximal with respect to the magnitude of the two initial numbers.
Applying the algorithm to two Fibonacci numbers $F {n}\,\, $ and $F_{n+1}\,\, $, the initial step is
$$ \gcd(F_{n},F_{n+1}) = \gcd(F_{n},F_{n+1}-F_{n}) = \gcd(F_{n-1},F_{n}) $$
The second step is
$$ \gcd(F_{n-1},F_{n}) = \gcd(F_{n-1},F_{n}-F_{n-1}) = \gcd(F_{n-2},F_{n-1}) $$
and so on. Proceding in this way, we need $n $ steps to arrive to $\gcd(F_{1},F_{2}) \,\,$ and to conclude that
$$ \gcd(F_{n},F_{n+1}) = \gcd(F_1,F_2) = 1$$
that is to say, two consecutive Fibonacci numbers are necessarily coprime.
Now it is well known that the growth rate of Fibonacci numbers, with respect to $n $, is exponential. In particular, $ F_n$ is asymptotic to
$${\displaystyle \varphi ^{n}/{\sqrt {5}}} $$
where
$$\varphi=\frac {1+\sqrt {5}}{2} \approx 1.61803$$
is the golden ratio. So, for $n $ sufficiently large, we have
$$n \approx \log_{\varphi} ( \sqrt {5}\,\, F_n) \\ = \log_{\varphi} ( F_n) + \frac {\log (5)}{2 \log (\varphi)} \approx \log_{\varphi} (F_n) $$
which tells us that the number of steps required to complete the Euclidean algorithm grows in a logarithmic manner with respect to $F_n\,\, $, and is bounded by the expression above. The same result can also be written using the logarithm in base $2$ as suggested by the OP:
$$n \approx \frac { \log_{2} F_n}{\log_{2} \varphi} =K \log_{2} F_n $$
where $$K=\frac {1}{\log_{2} \varphi}=\frac {\log {2}}{\log {\varphi}} \approx 1.4404... $$
Best Answer
I will address your first question.
While I think that the question is poorly phrased, I think I know what they mean by "What squares could you use to fill up a 6765$\times$4181 rectangle". I will illustrate this on a scaled-down version of the problem.
We have that $F_6=8$ and $F_5=5$. Let's fill up a $8\times5$ rectangle with squares.
One can see that we can fill up the rectangle perfectly with squares whose sides' lengths are all Fibonacci numbers up to $F_5$.
Why does this work for every pair of consecutive Fibonacci numbers?
Imagine that you are starting from one corner (e.g. the top right one as in the image) of the rectangle by placing a $1\times1$ square in it. Then, along the smaller side of the rectangle, place another $1\times1$ square. The two squares form another rectangle.
Now place another square ($2\times2$) next to the longer side of that rectangle. What results is yet another rectangle. Now just repeat the process until you use up all the Fibonacci numbers up to $F_6$.
Notice that by adding new squares, we will always get a new rectangle. What I mean by this is that the resulting figure will never be "jagged". This follows from the way that Fibonacci numbers are defined.
I think that this is what the question was going for, although it was not very well phrased, since there are many ways to fill up a rectangle with squares.
EDIT: The second part of the question
As Jair Taylor suggested in a comment, the number $F_{2018}/F_{2017}$ is rational and therefore its decimal representation will either have a finite number of digits or it will be periodic. This means that the decimal digits will contain repeating "chunks" of numbers. For example, in the decimal representation of the number $70/111$, the sequence "$630$" keeps repeating: $$70/111=0.\underbrace{630}\underbrace{630}\underbrace{630}...$$
Assuming that $\phi$ is the golden ratio, $\phi=\frac{1+\sqrt5}{2}$, its decimal representation will be unpredictable, because $\phi$ is an irrational number.
Looking at the pictures you provided, one can see that the first one is much more "chaotic" then the second one and one can hardly find any repeating patterns, whereas the second one obviously contains repeating patterns. This gives us good reason to conclude that the first image represents $\phi$ and the second one represents $F_{2018}/F_{2017}$.