``` sql ​ ​ 1) Get users who attended some activity more than twice. ​ """ SELECT * FROM users; ​ user_id username 1 John Doe 2 Jane Don 3 Alice Jones 4 Lisa Romero ​ SELECT * FROM training_details; ​ user_training_id user_id training_id training_date 1 1 1 "2015-08-02" 2 2 1 "2015-08-03" 3 3 2 "2015-08-02" 4 4 2 "2015-08-04" 5 2 2 "2015-08-03" 6 1 1 "2015-08-02" 7 3 2 "2015-08-04" 8 4 3 "2015-08-03" 9 1 4 "2015-08-03" 10 3 1 "2015-08-02" 11 4 2 "2015-08-04" 12 3 2 "2015-08-02" 13 1 1 "2015-08-02" 14 4 3 "2015-08-03" """ -------------------- ``` ``` python ​ # 2) Implement a group_by_owners that: # Accepts a dictionary containing the file owner name for each file name. # Returns a dictionary containing a list of file names for each owner name, in any order. # For example, dictionary: { 'Input.txt': 'Randy', 'Code.py': 'Stan', 'Output.txt': 'Randy' } ​ # the group_by_owners function should return: { 'Randy': ['Input.txt', 'Output.txt'], 'Stan': ['Code.py'] } # 3) Sort list of objects by age field in ascending order # Input: ​ animals = [ {'type': 'penguin', 'name': 'Stephanie', 'age': 8}, {'type': 'elephant', 'name': 'Devon', 'age': 3}, {'type': 'puma', 'name': 'Moe', 'age': 5}, ] ​ # Output: ​ sorted_animals = [ {'type': 'elephant', 'name': 'Devon', 'age': 3}, {'type': 'puma', 'name': 'Moe', 'age': 5}, {'type': 'penguin', 'name': 'Stephanie', 'age': 8} ] # 4) What will be the output of the following code snippet? class Person: def __init__(self, id): self.id = id ​ sam = Person(100) sam.__dict__['age'] = 49 print(sam.age + len(sam.__dict__)) ​ ​ # 5) What will be the output of the following code snippet? ​ class A: def __init__(self): self.calcI(30) print("i from A is", self.i) def calcI(self, i): self.i = 2 * i class B(A): def __init__(self): super().__init__() def calcI(self, i): self.i = 3 * i b = B() # 6) Write a code snippet that returns all pairs of numbers that return sum of 10 [0, -1, 9, 9, 1, 3, 6, 7, 1] ```