How to calculate the shannons channel capacity and the bandwidth, for a desired BER and EbNo value?
First, the Shannon formula tells you, given a certain amount of bandwidth and SNR, what is the maximum possible data throughput you can achieve. It does not tell you that any particular system or type of system can actually achieve that maximum. 

It doesn't tell you how to design a system --- it only tells you, after you designed it, how close you came to the theoretical maximum channel capacity.

Second, given an acceptable BER, you can calculate the (theoretical maximum) achievable bit rate from the channel capacity as 

\$R(p_b) = \dfrac{C}{1-H_2(p_b)}\$

where R(p) is the data rate, pb is the BER, C is the channel capcity, and H2(p) is the entropy function,

\$H_2(p_b)=- \left[ p_b \log_2 {p_b} + (1-p_b) \log_2 ({1-p_b}) \right]\$

So the achievable signal rate with 10-5 is about 1.0002 C. 

To answer your specific questions, the bandwidth in the Shannon formula is the minimum width of a perfect boxcar filter that could be applied to your signal without changing the signal. In the case of BFSK, it is very nearly \$\Delta{}f\$, the difference between the mark and space frequencies. As you say, for BFSK with the minimum modulation index, BW = 2 R, where R is the bit rate.

And the SNR in the Shannon formula is the same as the \$\frac{E_b}{N_o}\$ of your first formulation.

You have probably noticed that for this SNR and bandwidth (2*R), Shannon predicts a capacity of 

\$2 R \log_2\left(10^{1.2}\right)\$,

or 7.97 R, while your  BFSK is only achieving  R bit rate. (Note, there could be different defitinitions of the SNR that differ by a factor of 2 or so, but this only changes the result by +/- 1x(BW) after taking the log)

This just means that BFSK is only about 25% efficient in achieving the maximum channel capacity.