1.
hotel=# SELECT * FROM guest LEFT OUTER JOIN booking ON guest.phone_number = booking.guest WHERE booking.guest IS NULL;
 first_name | last_name | phone_number |         email         |               address               | check_in | check_out | guest | room
------------+-----------+--------------+-----------------------+-------------------------------------+----------+-----------+-------+------
 Marcus     | Cruz      |   6304468851 | rgarcia@optonline.net | 103 Water Ave. Grand Forks ND 58201 |          |           |       |
 Thalia     | Cobb      |   2269062721 | mxiao@yahoo.com       | 8929 Hall Court Flushing NY 11354   |          |           |       |
(2 rows)

                                                  ^
hotel=# SELECT guest.first_name, guest.last_name, booking.check_in, COUNT(booking.check_in) FROM guest JOIN booking ON guest.phone_number = booking.guest GROUP BY guest.first_name, guest.last_name, booking.check_in HAVING COUNT(*) > 1;
 first_name | last_name |  check_in  | count
------------+-----------+------------+-------
 Waylond    | Dalton    | 2017-10-22 |     3
(1 row)


hotel=# SELECT * FROM booking ou JOIN guest ON ou.guest = guest.phone_number  WHERE(SELECT count(*) from booking inr WHERE inr.room = ou.room AND inr.guest = ou.guest) > 1;
  check_in  | check_out  |   guest    | room | first_name | last_name | phone_number |        email        |                address
------------+------------+------------+------+------------+-----------+--------------+---------------------+---------------------------------------
 2017-11-10 | 2017-11-11 | 1255464478 |    9 | Justine    | Henderson |   1255464478 | fangorn@hotmail.com | 7591 Pheasant St. Schaumburg IL 60193
 2017-11-13 | 2017-11-14 | 1255464478 |    9 | Justine    | Henderson |   1255464478 | fangorn@hotmail.com | 7591 Pheasant St. Schaumburg IL 60193
 2017-11-15 | 2017-11-16 | 1255464478 |    9 | Justine    | Henderson |   1255464478 | fangorn@hotmail.com | 7591 Pheasant St. Schaumburg IL 60193
(3 rows)
2.
I will use three tables, one for students one for grades and one for classes.
Student table will hold the name, id, phone number address and other personal info on each student.
The class table will hold the class id, the subject, and the teacher assigned to it.
The grades table will have the student id, the class id and the grade earned.

3.
school=# SELECT * FROM student JOIN grades ON student.id = grades.student_id JOIN class ON grades.class_id = class.id WHERE class.id = 1;
           name            | id | student_id | class_id | grade | id | subject
---------------------------+----+------------+----------+-------+----+----------
 Waylon Dalton             |  1 |          1 |        1 | A     |  1 | Language
 Justine Henderson         |  2 |          2 |        1 | F     |  1 | Language
 Marcus Cruz               |  4 |          4 |        1 | B     |  1 | Language
 Thalia Cobb               |  5 |          5 |        1 | C     |  1 | Language
(4 rows)

school=# SELECT grade FROM grades WHERE class_id = 1 ORDER BY grade;
 grade
-------
 A
 B
 C
 F
(4 rows)

school=# SELECT class.subject, COUNT(grades.student_id) FROM grades JOIN class ON grades.class_id = class.id GROUP BY class.subject;
     subject      | count
------------------+-------
 Computer Science |     4
 History          |     4
 Science          |     3
 Math             |     5
 Language         |     4
(5 rows)

school=# SELECT class.subject, COUNT(grades.student_id) FROM grades JOIN class ON grades.class_id = class.id GROUP BY class.subject ORDER BY count DESC LIMIT 1;
 subject | count
---------+-------
 Math    |     5
(1 row)