sinus generation and it's fft in radians
Why on first plot fft shows that cosine has a frequency 11 radians?
  


The array index number of the cosine peak is 11.  The frequency is different.  

Matlab indexes from 1, so the FFT bin number will be 10 (as we have a 0 bin in FFTs).

The frequency of that bin will be 
$$
\frac{2 \pi (11 - 1) }{N } = \frac{2 \pi 10}{ 198} = \frac{10\pi}{99}
$$
assuming your FFT length was 198.

Note that this will be a normalized frequency between 0 and $2\pi$ (or, if you prefer, $-\pi$ and +$\pi$).


  
  How to shift fft to achieve frequency domain around 0?
  


Your use of fftshift does this in the second plot.  However, you need to set the $x$ axis by doing something like:

N = length(sinefft);
delta_omega = 2*pi/N; % rad/sec between FFT bins
omega = -pi:delta_omega:(pi - delta_omega)
plot(omega, abs(fftshift(sinefft)));


Minor Correction I previously had this frequency variable as f, it's actually omega.


  How to calculate now what frequency in Hz has this signal?


So the normalized frequency in radian per second is omega.  That ranges from $-\pi$ to $+\pi$ (modulo some end effects).

The normalized frequency in "Hz" is f. This will be omega/2/pi. This ranges from -0.5 to +0.5.

The frequency in terms of the sampling frequency will be f_true = omega / pi / 2 * fs (where fs is the sampling frequency).