Spectroscopy pyrazine molecule
You need to understand some details of using point group tables. The pyrazine  belongs to  $\ce{D_{2h}}$ point group and pyridazine belongs to $\ce{C_{2v}}$
In the third column of the point group are found x, y, z which are used to identify dipole transitions, e.g. vibrational symmetries that can exhibit  IR transitions and in the fourth $x^2, z^2, xz$ etc. which are used to identify Raman transitions. If there is a squared term in column 4 and in the top line of the table (totally symmetric representation) labelled $\ce{A_g, A_1}$ then a Raman transition is totally polarised. Raman transitions from other symmetry species vibrations are not fully polarised. (Ignore the $\ce{R_x, R_y,R_z}$ they do not refer to IR or Raman transitions)
The point groups are shown below
  
In $\ce{C_{2v}}$ all vibrational species $\ce{A_1}$ etc. can have Raman transitions but in the IR the $A_2$ does not have an IR transition, no x, x, z in column 3. The other three types of vibrations could have both IR and Raman transitions. In the $\ce{D_{2h}}$ you can see that, because of the centre of inversion, Raman and IR vibrational transitions are mutually exclusive (compare symmetry species of $x^2, xy$ etc. with x, y, z ) because there is no vibration with a symmetry species that can generate both IR and Raman transitions.
Thus, with your compounds, the one with $\ce{C_{2v}}$ symmetry will have IR and Raman transitions with the same frequency, but for the $\ce{D_{2h}}$ compound there should be no common IR and Raman frequencies.
(See the answer to this question Understanding group theory easily and quickly more details on using point groups)