Product of the reaction between chalcone and sodium methoxide in methanol
The first step is called Michael addition. Have a look at this wikipedia page.


  The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated carbonyl compound.


Of course, in your case, the nucleophile is $\ce{OCH3-}$. Perform 1,4-addition by adding it to the double bond at the beta position (with respect to the carbonyl), and you'll expect an enolate ion, which looks like this - 


  


Finally protonate the $\ce{O-}$ and tautomerise to get your product as @Zenix already stated. 

For the second step, check out this link. 


  


References: Michael reaction, wikipedia || Ketone halogenation, wikipedia || Enol, wikipedia