truth tables

b1b2 !b1!b2 b1 && b2 b1 || b2
truetrue false false true true
truefalse false true false true
falsetrue true false falsetrue
falsefalse true true false false

DeMorgans

!b1 || ! b2!b1 && !b2!(!b1 || ! b2)!(!b1 && !b2)
false false true true
true false false true
true false false true
true true false false

3.

(i != 1 || i != 0)
first think about it and determine what you think the answer is

Negation should give us everything other than the solution we thought of

negation would be

! (i != 1 || i != 0)
by deMorgans ! (i != 1) && ! (i != 0)

i = 1 && i = 0

never, ever true. Hence answer above should have been always true.