Question 1046573
1.  (P v Q)=>R => (P=>R)^(Q=>R)
Proof:  
 (P v Q)=>R         
---> ~(P v Q) V R    (Material implication)
---> (~P ^ ~Q) v R   (de Morgan's)
---> (~P v R) ^ (~Q v R)   (distributivity)
---> (P=>Q) ^ (Q=>R)    (Material implication)

To prove  (P=>R)^(Q=>R) => (P v Q)=>R, just reverse the steps in the preceding argument.


2. By addition, it is easy to see that P => P v (P^Q) .
To prove P v (P^Q) => P:

P v (P^Q)
---> (PvP)^(PvQ)  (distributivity)

---> P^(PvQ)    (idempotency)
---> P       (simplification)

Therefore, P v (P^Q) = P.


3.  I leave this up to you.