Solutions (Pseudocode):
1. void FizzBuzz(int n){
     for(int i = 0; i < n; i++){
     if(i % 3 == 0)
       print "Fizz"
     if(i % 5 == 0)
       print "Buzz"
     print "\n"
}

2: Returns true when n is a power of two
3: C++ Solution int HasCycle(Node* head)
{
    Node *n1 = head;
    Node *n2 = head;
//make sure neither are null, or we don't have a cycle
    while(n1 && n2){
//2 iterators for the list
      n1 = n1->next;
      n2 = n2->next;
      if(!n2)
        return 0;
      else
        n2 = n2->next; //n2 goes "twice as fast" as n1
     //this is how we know there is a cycle
      if(n1 == n2)
        return 1;
    }
    return 0;
}

4: C++ Solution:

void Inorder(node *root) {
    if(!root) { return; }
    Inorder(root->left);
    std::cout << root->data << " ";
    Inorder(root->right);
}


5: Java Solution:
Node MergeLists(Node list1, Node list2) {
  if (list1 == null) return list2;
  if (list2 == null) return list1;
  if (list1.data < list2.data) {
    list1.next = MergeLists(list1.next, list2);
    return list1;
  } else {
    list2.next = MergeLists(list2.next, list1);
    return list2;
  } }

6: C++ Solution
void ReversePrint(Node *head)
{
    if(head->next)
      ReversePrint(head->next);
    std::cout << head->data << std::endl;
}

7: Python solution def find_intersection(values):
    all_uniques = []
    for value in values:
      all_uniques.append(set([l for l in value]))
    return len(reduce(lambda x, y: x & y all_uniques))


8: Solution TBD