The problem :
My issue : I tried this problem in a lot many ways, but none worked out, I tried finding a way, where it would kind of form any pattern and thus will be easy to multiply all terms or in some way cancel out, but none materialised. (I am not showing here any of my trials for they are too trivial, to lead anywhere),
so can you help me here any way ?
We can use Binet's formula, which states that $$F_n=\frac{1}{\sqrt{5}}\left(\phi^n - \left(-\frac{1}{\phi}\right)^n\right)$$ for all $n$ where $\phi=\frac{1+\sqrt{5}}{2}$
Consider $$\prod_{k=2}^\infty \frac{F_{2n}-1}{F_{2n}+1}$$
Using Binet's formula, the terms in the product become $$\frac{F_{2n}-1}{F_{2n}+1}=\frac{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)-1}{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)+1}=\frac{\phi^{4n}-\sqrt{5}\phi^{2n}-1}{\phi^{4n}+\sqrt{5}\phi^{2n}-1}$$
Noting that $\phi^2-\frac{1}{\phi^2}=\sqrt{5}$, we see that this can be factorised as $$\frac{\left(\phi^{2n}-\phi^2\right)\left(\phi^{2n}+\phi^{-2}\right)}{\left(\phi^{2n}-\phi^{-2}\right)\left(\phi^{2n}+\phi^2\right)} = \frac{\left(\phi^{2n-2}-1\right)\left(\phi^{2n+2}+1\right)}{\left(\phi^{2n+2}-1\right)\left(\phi^{2n-2}+1\right)}=\frac{a_{n-1}b_{n+1}}{a_{n+1}b_{n-1}}$$ where we define $$a_n=\phi^{2n}-1$$ and $$b_n=\phi^{2n}+1$$
The product then becomes $$\prod_{k=2}^\infty \frac{a_{k-1}b_{k+1}}{a_{k+1}b_{k-1}}$$ which telescopes, and we see that the product is equal to $$\lim_{n\to\infty} \frac{a_1 a_2 b_n b_{n+1}}{a_n a_{n+1} b_1 b_2}$$
You can check that $$\lim_{n\to\infty} \frac{b_n}{a_n} = 1$$ so that the product simplifies to
$$\frac{a_1 a_2}{b_1 b_2}=\frac{(\phi^2-1)(\phi^4-1)}{(\phi^2+1)(\phi^4+1)}=\frac{(\phi^2-1)^2}{\phi^4+1}=\frac{\phi^2}{3\phi+2+1}=\frac{1}{3}$$
(Since $\phi^2=\phi+1$, and so $\phi^4=\phi^2+2\phi+1=3\phi+2$)
So I guess your answer would be $3$.