For point 1: Yes, you are absolutely right. And this is justified by just another law of Boolean algebra, namely a∧a = a which you can, of course, also apply the other way around, i.e., a = a∧a.

For point 2: As Martin pointed out, you can add first redundant clause, using the mentioned simplification law from the other direction, i.e., a = a∧b ∨ a∧¬b. This works as follows:

a ∨ ¬a∧c ∨ ¬a∧¬b

= a ∨ a ∨ ¬a∧c ∨ ¬a∧¬b

= a∧c ∨ a∧¬c ∨ a∧b ∨ a∧¬b ∨ ¬a∧c ∨ ¬a∧¬b

= (a∧c ∨ a∧¬c) ∨ (a∧c ∨ ¬a∧c) ∨ (a∧¬b ∨ ¬a∧¬b) ∨ a∧b

= a ∨ c ∨ ¬b ∨ a∧b

= a ∨ c ∨ ¬b

where yet another law was used, namely a ∨ a∧b = a which is the absorption law of boolean algebra.