Weak bisimulation up-to $\approx$
(partial answer)


  
  For the relation $\mathcal{R} = \{(\tau.a, 0)\}$, $\mathcal{R} \nsubseteq \approx$ but $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$. Why is $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$? 
  


Actually, the inclusion $\mathcal{R} \subseteq \, \approx \! \mathcal{R} \approx$ holds for any relation $\mathcal{R}$, and not just the one mentioned. Indeed, if we have $P \mathcal{R} Q$, then we also have $P \approx P \mathcal{R} Q \approx Q$, by reflexivity of weak bisimulation.


  
  According to Sangiorgi's book, P uses $\tau$ and ends up with $a \approx \tau.a$ and Q has $0$ and ignores the move. I am wondering, how is it that you can ignore the move? I thought that one is forced to do a move...
  


You have to check that $0\xrightarrow{{\hat\tau}} 0$, which (if I remember correctly) is defined as $0\xrightarrow{{\tau}}^* 0$ which holds since the Kleene's star includes "zero steps".

I seem to remember that the "$\hat\alpha$" move definition made a special case for $\hat\tau$, specifically to allow for zero steps.