Predict the product (organic chemistry)
Looks like a classic to me. But since this is homework, I will not provide a full reaction scheme and only give some hints.


Your starting material is a $\beta$-ketoester (1,3-dicarbonyl compound).


Where is the most CH-acidic position in this molecule?
Benzylchloride is an alkylating agent. What will happen when it reacts with the anion of your starting material?
For extra credit: Why is sodium ethoxide in ethanol a reasonable choice when you want to isolate, purify and characterize the product? 

Your starting material did have an ester group. Does it still have it after the first step? If so, what will happen to it in aqueous alkaline solution? 
(This step would not work with the saponification product of a simple ester.) 
But you started with $\beta$-ketoester! Heating the product of step 2 in acidic media will let it loose some atoms. The mechanism may be drawn using a cyclic intermediate, in which the ketone has an assisting role.



EDIT


Deprotonation and subsequent benzylation of the $\beta$-ketoester occur at the most CH-acidic position, i.e beween the keto and the ester group. As a side note, benzylation at the methyl group adjacent to the carbonyl group is possible when using 2 equivalents of a strong base.

Using ethoxide rather than methoxide as a base makes sure that transesterification does not lead to a mixture of products.

The resulting benzylated $\beta$-ketoester is cleaved under alkaline conditions to the corresponding carboxylate. Under acidic conditions, this from a $\beta$-ketoacid.

Upon heating, $\beta$-ketoacids undergo decarboxylation (= they loose $\ce{CO2}$) to form an enolate which tautomerizes to the final methyl ketone.