The hydride(Strong base) will attack the alpha H and become H2.Electron will attack the carbonyl carbon(Since there is no other electrophilic molecule) and cause the elimination of -OC2H5.
Your products are correct, but your reasoning is questionable.
You do not have free hydride here, DIBAL is not a source of basic hydride (whereas NaH is, for example). DIBAL itsself is a lewis acidic reducing agent.
Collapse of the tetrahedral intermediate, eliminating ethanol, occurs during workup, not directly following reduction - this is the root of the reason why DIBAL will normally reduce esters only to aldehydes and not all the way to the alcohol.
You ought to look up the mechanism of this reduction, the wikipedia page is very brief and only meant as an introduction, any university level organic chemistry text will do, eg. Clayden
Your answer to 2 is wrong,
The C2H5O-(act as base) attack the alpha hydrogen of CH3COOC2H5.Then the CH2(-)COOC2H5 will attack hydroxyl C of CH3CH2OH and give out -OH.
You are correct in your generation of the enolate, but this enolate will not attack ethanol to displace hydroxide as you have described. Hydroxide is a very poor leaving group. Are there any more electrophilic compounds in your solution..?
If you read the wiki page on the Claisen condensation more carefully you will see that it contains the answer for this exact question, and the mechanism.
I strongly advise you to sit down with a good textbook and get a clearer notion of what's going on. If you're still unsure you will certainly get more help here.