Converting to CNF - Distribution
By definition - for example Herbert B. Enderton - A Mathematical Introduction to Logic-Academic Press (2001), page 53 - simply saying CNF is conjuction of disjuntions or product of sums.

So, as we agreed earlier,
$$(I \lor M) \Rightarrow H = (\neg I \lor H) \land (\neg M \lor H )$$

For $(I \land M) \Rightarrow H$ we can get $\neg (I \land M) \lor H=\neg I \lor \neg M \lor H$ and this last is CNF

For third case for $(\neg I \land \neg M) \Rightarrow H$ answer is analogical $2$d case: $\neg(\neg I \land \neg M) \lor H = I \lor M \lor H$ and this also is valid CNF.


And answer for the last question is yes - $A_1 \lor A_2 \lor \cdots  \lor A_n$ is valid CNF if each $A_i$ is  either a sentence symbol or the negation of a sentence
symbol - it's allowed to have just one conjuct.
Let me add, that one symbol is as CNF, so DNF.