For part $(1)$, Since $\frac{|\phi|^n}{\sqrt 5}\lt \frac{1}{2}$ $F_n=\lfloor \frac{\phi^n}{\sqrt 5}+\frac{1}{2}\rfloor$ as where $\lfloor . \rfloor $ represents the greatest integer function
For part $(3)$, For even $n$,$F_n=\frac{\phi^n-\frac{1}{\phi^n}}{\sqrt 5}=\frac{\phi^{2n}-1}{\phi^n\sqrt 5}$ $\implies (\phi^n)^2-(\sqrt 5 F_n) (\phi^n)-1=0$.
This is quadratic equation in $\phi^n$ solving which we get,
$\phi^n=\frac{\sqrt 5F_n+\sqrt{5F_n^2+4}}{2}$ (negative root discarded) $\implies n=\log_\phi \frac{\sqrt 5F_n+\sqrt{5F_n^2+4}}{2}=\frac{\ln (\frac{\sqrt 5F_n+\sqrt{5F_n^2+4}}{2})}{\ln \phi}$
As you can see that for large $F_n$, $\sqrt{5F_n^2+4}\approx\sqrt 5 F_n$ which gives $n\approx\frac{\ln \sqrt 5F_n}{\ln \phi}$ for large $F_n$
For odd $n$, similar computations follow which give
$n=\frac{\ln (\frac{\sqrt 5F_n+\sqrt{5F_n^2-4}}{2})}{\ln \phi}$
But since you don't know whether $n$ is even or odd,
my suggestion is to compute both expressions (for even one as well as odd one) and check which $n$ fits best.
For large $F_n$, those expression would yield same $n$, so no need to worry for large $F_n$