Instead of a grouped COUNT you can use it as a windowed aggregate to access the other columns

SELECT fullname,
       address,
       city
FROM   (SELECT *,
               COUNT(*) OVER (PARTITION BY fullname, city) AS cnt
        FROM   employee) e
WHERE  cnt > 1

Agree with above answer. But If you don't want to use Windows functions which might not work properly on all DBs you can join to itself on city and full name after the group by and having and then get the addresses

Select employee.* from employee 
   join (select fullname, city from employee group by fullname, city having count(*)>1) q1 
   on q1.fullname = employee.fullname and q1.city = employee.city

Try the Following Code:

create table ##Employee
      (Fullname varchar(25),
       Address varchar(25),
       City varchar(25))

      insert into ##Employee values
     (    'AA',          'address1',    'City1')
    ,(    'AA',          'address3',    'City1')
    ,(    'AA',          'address8',    'City2')
    ,(    'BB',          'address5',    'City2')
    ,(    'BB',          'address2',    'City1')
    ,(    'CC',          'address6',    'City1')
    ,(    'CC',          'address7',    'City2')

      select E.* from ##Employee E
      cross apply(
      select Fullname,City,count(Fullname) cnt from ##Employee
      group by Fullname,City
      having Count(Fullname)>1)x
      where E.Fullname=x.Fullname
      and E.City=x.City

If you have a unique id or the address is always different, you can try:

select e.*
from employee e
where exists (select 1
              from employee e2
              where e2.fullname = e.fullname and e2.city = e.city and
                    e2.address <> e.address  -- or id or some other unique column
             );

Although I would probably go with the window function approach, you might find that under some circumstances, this is faster (especially if you have an index on employee(fullname, city, address) ).

Here you go with the solution:

DECLARE  @Employee TABLE
        (
            Fullname VARCHAR(25),
            [Address] VARCHAR(25),
            City VARCHAR(25)
        )

      INSERT INTO @Employee VALUES
      ('AA', 'address1', 'City1') 
      ,('AA', 'address1', 'City1') 
      ,('AA', 'address3', 'City1')
      ,('AA', 'address8', 'City2')
      ,('BB', 'address5', 'City2')
      ,('BB', 'address2', 'City1')
      ,('CC', 'address6', 'City1')
      ,('CC', 'address7', 'City2')

     ;WITH cte AS (
               SELECT *,
                      ROW_NUMBER() OVER(PARTITION BY FullName, [Address], [City] ORDER BY Fullname) AS sl,
                      HashBytes('MD5', FullName + [Address] + [City]) AS RecordId
               FROM   @Employee AS e
           )

      SELECT c.FullName,
             c.[Address],
             c.City
      FROM   cte             AS c
             INNER JOIN cte  AS c1
                  ON  c.RecordId = c1.RecordId
      WHERE  c.sl = 2

Result :

FullName    Address     City
AA          address1    City1
AA          address1    City1

SELECT Feild1, Feild2, COUNT() FROM table name GROUP BY Feild1, Feild2 HAVING COUNT()>1

This will give you all yours answer