a)The third smallest element can occupy positions 2,3,4,5,6,7
because the smallest element being the root occupies position 1.

If in position 2 is the second smallest element, 
the third smallest element can be in position 3 because there is no contraints between siblings. 

If second smallest is in position 3, third smallest can be in position 2.

If the second smallest element is in position 2 then the third smallest can be in positions 4 and 5 (children of 2) because the children of 2 are not constrained by the sibling of 2.

If the second smallest element is in position 3 then the third smallest can be in positions 6 and 7 (children of 3) because the children of  3 ar enot constrained by the sibling of 3


b)
To maintain the min-heap property the largest element can only be in a position where it does not have any children larger than it meaning it can only be a leaf Positions ( (ceil(n/2) + 1).........n)


c) The rank n-1 element , second largest element could be in a leaf position because then then it would be a sibling of the largest element and there no conraints defining siblings.
Also it could be a parent of 1 child (largest element) only.
Because if the node had 2 children then it means it has 2 elements larger than it and cannot be of rank n-1

d)

1: When n is odd there are (n-1)/2 elements greater than the median and (n-1)/2 elements lesser than the median. 
   Node positions that have no more than (n-1)/2 descendants and no more than (n-1)/2 ancestors are candidates for the 
   median positions.
   Beyond position 2 the node position with the most number of children is position 3. A node in position 3 has atmost         ((n-1)/2)-1 descendants. All the other positions for nodes have lesser number of descendants. 
   The nodes with most ancestors are the nodes in the leaf positions. Since the height of the tree is lg(n) they have 
   lg(n)-1 ancestors. 
   
    so we have ((n-1)/2) < (n-1)/2
    	       lg(n)-1   < (n-1)/2 
   This means any position beyond position 2 is a suitable position for the median.


3:  n= 2^j-4 .........2^j+2
	For the upper median
    n= 2^j-3..........2^j+1
	For the lower median