You should do your homework. If you would liste to your teacher or read your course note, it would not be that hard at least for some cases...
For case (i), you could you could distribute
B || C
to the rest of the equation and then you can simplify the equation as one part is included in the other.
Case (ii) is very trivial to solve.
Case (iii) is somewhat more difficult but you use rule like
!X || !Y
is equivalent to
!(X && Y)
a few times. Alternatively, you can draw a truth table, and find the solution.
A B C !A !C (...) C !B (...) result
0 0 0 1 1 1 0 1 1 1
0 0 1 1 0 1 0 1 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Case (iv) has some trivial simplification...
You have to do other exercise all by yourself... So later cases seems to be incorrect either as code or mathematical expression. Also, for some of those cases, other information is required...