Name the alkene belo. Use °NINE. designators to indicate stereoche.stry. 
H H HaC \/C= =C C CI-13 CH3CH2C/F12 CH3 
 

You can do this with Cahn-Ingold-Prelog rule. This rule gives priority to the ramifications of the alkene.

Atoms with higher atomic number has priority, if there is (like in these case) a tie, we must sum the atomic number of the next atom of the ramification.

So in this case

The ramifications that are in a red square are the most important, has the higher priority, because if we sum the atomic number of the atoms we get the higher results. One of the ramifications is "above" the doble bond and the another is "below", so, in these case we call it E. We have to do the same with the another doble bond, in this case we see that it is Z because the higher priority ramifications are "below" the doble bond.

The molecule is an octadiene, because it has 2 doble bonds and 8 C atoms. Then we have 2 metil ramifications, lets draw the molecule again to see the ramifications:

The name of the alkene will be: (2Z,4E)-4,5-dimethyl-2,4-octadiene

please see the attached file for the complete solution