Is (A ∨ B) → (A ∧ B) possible? Hi still fairly new to propositional logic, and I'm attempting to prove 
(A ∨ B) → (A ∧ B)

I have tried via De Morgan Laws but to no success.

If not, would this work?
A → (A ∧ B)

Any help will be appreciated.
You can't prove it because it isn't true.  If $A$ is true and $B$ is false, $A\vee B$ is true and $A \wedge B$ is false, so the whole statement is false.  Just read it in English as (A or B) implies (A and B) and it doesn't make sense that it would be true.
That proposition that you are trying to prove is NOT true: $$(A\lor B) \to (A\land B)\;\;\color{red}{\Huge \checkmark}$$

When we have the antecedent $A\lor B$, it is true when $A$ is true, but not $B$, or when $B$  is true, but not $A$.  However, in either of those cases, $A\land B$ is not true, because $A \land B$ is true only when both A and B are true.  For example, If I say "I'll have pie, or cake for desert", I will not be lying if I have either pie or cake, but not both.  However, if that's my promise, no one can accuse me of claiming therefore "I'll have pie AND cake for desert."

Whenever the antecedent can be true, but the consequent false, the implication is fails.



For your second question, it is not true that $A\to (A\land B).$  If A is true, but B is false, then the antecedent is true, but the consequent is false, because (true $\land$ false$ evaluates to false. 

What is true is that $$[(A\land B) \to A] \land [(A\land B) \to B]$$

What is true is that $(A\land B) \to (A\lor B)$.