i don't know what kind of logic/proofs you are studying, so i don't know how you want these set up, but here's the idea for this one.

clearly s or ~s is true. if s is true, then we have ~a & b. however, this is the negation of a or ~b (DeMorgan's law), and so we cannot have that. thus, ~s must be true. but ~s => c, so that we can conclude c

i don't think this is correct. are you sure there is no typo?2) ~r

( b or c) -> r

~d or m

d -> ~ ( c or p)

_______________

p -> m

we must show either s or q is true, perhaps both. since we have b => c and ~c, then we have ~b. if ~b is true, then ~b or ~p is true. and hence, the last implication says s is true. if s is true, then s or q is true, and we conclude accordingly.3) b -> c

~c

(~b or ~p) -> s

_______________

s V q

i leave the formal proofs to you