Reduction of alkyl halides to alkanes
Your answer for question #3 is correct. However, there is a fundamental misconception you used in #1 and #2 that led to an incorrect answer.

In your notes, it seems that electrons from the metal ($\ce{Zn}$ and $\ce{Mg}$) somehow "attack" the alkyl halide and generate the methine ion ($\ce{CH3-}$). This does not happen for two reasons: Firstly, electrons are not free species in solution (like ions and molecules), they are only transferred from one molecule (reductant) to another (oxidant), so they cannot attack anything by themselves. Secondly, the methine ion is an extremely strong base (the pKa for methane is around 50, so the methine ion, its conjugate base, has to be VERY basic), which could only be produced by using an even stronger base to abstract a proton from it. 

What actually happens in both cases is what is called metallic insertion. The metal ($\ce{Zn}$ or $\ce{Mg}$) is inserted between the carbon-halide bond, forcing the carbon atom to hold a partial negative charge (the author of this picture forgot to include the $\ce X$ atom bound to $\ce M$ in the organometal compound, but it is not that relevant here).



https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm

When it happens, you end up having a nucleophilic carbon, and it could indeed react with a $\ce{CH3Cl}$ molecule to yield $\ce{CH3-CH3}$ under a unimolecular substitution mechanism ($\ce{S_N1}$). However, such substitution reactions are very slow compared to acid-base reactions, and since you have acidic hydrogens available in both cases ($\ce{HCl}$ in question #1 and $\ce{H2O}$ in #2) the acid-base reaction (abstraction of a H atom from $\ce{HCl}$ or water) would happen fast enough to prevent the formation of $\ce{CH3-CH3}$ adduct.