Question 1: Master Theorem not possible since b=1, which is not greater than 1.
Question 2: Master Theorem not possible since b=1, which is not greater than 1.
Question 3:
T(n)=3T( (n/3) - 2 ) +n/2 = 3T((3n-6)/3)+n/2 = 3T(3(n-2)/3)+n/2 = 3T((n-2))+n/2
Now, Master Theorem not possible since b=1, which is not greater than 1.
Question 4: T(n)=T(n/2)+T(n/4)+T(n/8)+n
Roughly, T(n) can be written as T(n)<T(n/2)+T(n/2)+T(n/2)+n, which can be further reduced to
T(n)<3T(n/2) + n
Now, a = 3, b=2, k=1, p=0
Clearly, a > bk i.e. 3>21.
So, Case 1 satisfies. Therefore, T(n)=Theta(nlg3) = Theta(n1.584962...) = Theta(n)
I hope it helps. :)