I am trying to figure out if the infinite product $$\omega=\frac{5\sqrt{3}}{12}\prod\limits_{\substack{p\equiv 1\pmod3 \\ p\ge 13}}\left(\frac{p-2}{p-1}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\ p\ge 13}}\left(\frac{p}{p-1}\right)$$ is asymptotically equal to the infinite product $$c=\frac{5775}{2592\pi}\prod\limits_{\substack{p\equiv 1\pmod3 \\ p\ge 13}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\ p\ge 13}}\left(\frac{p^2}{p^2-1}\right)$$.
So, I reformulated the products as $$\omega(x)=\frac{5\sqrt{3}}{12}\prod\limits_{\substack{p\equiv 1\pmod3 \\ 13\leq p\leq x}}\left(\frac{p-2}{p-1}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\ 13\leq p\leq x}}\left(\frac{p}{p-1}\right)$$ and $$c(x)=\frac{5775}{2592\pi}\prod\limits_{\substack{p\equiv 1\pmod3 \\ 13\leq p\leq x}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\ 13\leq p\leq x}}\left(\frac{p^2}{p^2-1}\right)$$. Then, taking the quotient we get $$\frac{\omega(x)}{c(x)}=\frac{216\sqrt{3}\pi}{1155}\prod\limits_{\substack{p\equiv 1\pmod3 \\ 13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\ 13\leq p\leq x}}\left(1+\frac{1}{p}\right)$$ $$=\frac{216\sqrt{3}\pi}{1155}\prod\limits_{\substack{p\equiv 2\pmod3 \\ 13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)\prod\limits_{\substack{p\equiv 1\pmod3 \\ 13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits_{\substack{p\equiv 2\pmod3 \\13\leq p\leq x}}\left(\frac{1}{1-\frac{1}{p}}\right).$$
As $x\to\infty$, from A. Languasco's paper, we observe that the last two products in the quotient above are asymptotically equal. So, the limit of the quotient depends on the product $$\frac{216\sqrt{3}\pi}{1155}\prod\limits_{\substack{p\equiv 2\pmod3 \\ 13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)$$, but I don't know if this converges to $1$ or not as $x\to\infty$. I can see that the last product (except the constant) has something to do with $\frac{1}{\zeta(2)}$ but I don't know if it's actually smaller or greater than that.
I would really appreciate it if somebody could provide some hints or ideas on how to progress from here.
$\chi_3(3n)=0,\chi_3(3n+1)=1,\chi_3(3n+2)=-1$ is the Dirichlet character $\bmod 3$.
You need the PNT (or a Mertens theorem) for $L(s,\chi_3)=\sum_{n\ge 1} \chi_3(n)n^{-s}$ to show that both products converge.
No reason to start at $13$, I'll start at $2$.
Their quotient is $$\prod_p \frac1{1-\chi_3(p)p^{-1}}$$
Using $\chi_3(n)= \frac{e^{2i\pi n/3}-e^{-2i\pi n/3}}{\sqrt3 i}$,
by the PNT (or a Mertens theorem) for $L(s,\chi_3)$ this product converges to $$L(1,\chi_3) = \sum_{n=1}^\infty n^{-1} \frac{e^{2i\pi n/3}-e^{-2i\pi n/3}}{\sqrt3 i}= \frac{-\log(1-e^{2i\pi/3})+\log(1-e^{-2i\pi/3})}{\sqrt3 i}=\frac{\pi}{3\sqrt 3}$$