I just found that my question (2) is trivially easy and that indeed it is:

$$\begin{array}{l}

m = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - {a_B} - {{\tilde a}_{II}}} \right)} \\

{\rm{ }} = \sqrt {\frac{{2\pi T}}{{{c_0}}}\left( {B + {{\tilde N}_{II}} - 1 - {{\tilde a}_{II}}} \right)}

\end{array}$$

This is because:

$$m=\sqrt{\frac{2\pi T}{c_0}\left(N+\tilde N-a-\tilde a\right)}$$

***Edit: *** Ooops. I was wrong, The correct answer is (in the RR sector): $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N + \tilde N+2}} $$ In the N-SN-S sector, the correct answer is: $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N +\tilde N - 2}} $$ In the RN-S/N-SR sectors, $$m = \frac{2\pi T\ell_s}{c_0^2}\sqrt {{N+\tilde N}} $$

This is because its Ramond sector normal-ordering constant is $-1,$ while its Neveu-Schwarz sector normal-ordering constant is $1.$