I believe that's is possible on assumption of the Twin Prime Conjecture, that $$\sum_{n=1}^\infty\frac{\mu(n)\pi_2(n)}{n^2}$$ converges following the guideline of the answer (see the reference).
One has that there exists a integer $n_0$ such that being $\pi_2(x)$ the twin prime-counting function $$\sum_{n = 2}^\infty \frac{\mu(n)}{n}\biggl(\frac{\pi_2(n)}{n} - \frac{1}{\log^2 n}\biggr),$$ converges absolutely because on our assumption $$\frac{\pi_2(n)}{n} - \frac{1}{\log^2 n} \in O\bigl((\log n)^{-3}\bigr),$$ since if the Twin Prime conjecture is true then $\pi_2(n)=O\left(\frac{n}{\log ^2 n}\right)$.
And now, one has that since the sequence $\left(\frac{1}{\log ^2 n}\right)_{n\geq n_0}$ converges to $0$ and is monotonically decreasing, so by Dirchlet's test $$\sum_{n = n_0}^\infty \frac{\mu(n)}{n\log ^2 n}$$ converges (conditionally). It follows that $$\sum_{n = n_0}^\infty \frac{\mu(n)\pi_2(n)}{n^2}$$ converges (conditionally), as it is the sum of two convergent series.
Question. On assumption of the Twin Prime Conjecture was right my calculations? Can you improve or provide us some details of some step, especially doing focus about convergence issues (with our assumption)? Many thanks.
Reference:
This was from an answer of my previous post What's about the convergence of $\sum_{n=1}^\infty\frac{\mu(n)\pi(n)}{n^2}$?