Relational Algebra Queries - How to Approach Problems?
I think you're on the right track, but I'm not seeing a provision for restricting the results to employees who are actually working on two projects and the rename notation is confusing to me.  You'll probably want to leverage a theta-join on a self join to accomplish this.

To start out, I would rename the relevant attributes in the work table in preparation for a self-join:


  ρsin/emp_sin,project_number1/project_number works_on


Then project just those attributes for simplicity's sake:


  πsin,project_number1 ( ρsin/emp_sin,project_number1/project_number works_on ) 


For visualization purposes, the equivalent SQL here would be something along the lines of:

SELECT  emp_sin 
            AS sin,
        project_number
            AS project_number1
FROM    works_on;


I'd then prepare for the self join by creating another relationship, renaming the project_number differently:


  πsin,project_number2 ( ρsin/emp_sin,project_number2/project_number works_on ) 


At this point, a theta-join can be applied, exploiting the maximum of 2 carnality as defined in the assumption.  This theta-join creates a relationship with both project_numbers by emp_sin:


  ( πsin,project_number1 ( ρsin/emp_sin,project_number1/project_number works_on ) ) ⋈sin, project_number1 &lt; project_number2 ( πsin,project_number2 ( ρsin/emp_sin,project_number2/project_number works_on ) )


In effect, this is similar to SQL as follows:

SELECT  *
FROM (  SELECT  emp_sin 
                    AS sin,
                project_number
                    AS project_number1
        FROM    works_on ) AS w1
INNER JOIN (
        SELECT  emp_sin 
                    AS sin,
                project_number
                    AS project_number2
        FROM    works_on ) AS w2 
    ON  w1.sin = w2.sin
    AND w1.project_number1 &lt; w2.project_number2;


The little exploit in the requirements here lets us do a natural join and a final projection to finish up ( since I've already renamed the emp_sin attribute to sin ):


  πemp_name, project_number1, project_number2 ( employee ⋈ ( πsin,project_number1 ( ρsin/emp_sin,project_number1/project_number works_on ) ) ⋈sin, project_number1 &lt; project_number2 ( πsin,project_number2 ( ρsin/emp_sin,project_number2/project_number works_on ) ) )


Again, as visual proof, this is approximately the equivalent of the following SQL, shown in action in this SQL Fiddle:

SELECT  e.emp_name, w1.project_number1, w2.project_number2
FROM    dbo.employee e
INNER JOIN (
        SELECT  emp_sin 
                    AS sin,
                project_number
                    AS project_number1
        FROM    dbo.works_on ) AS w1
    ON  e.sin = w1.sin
INNER JOIN (
        SELECT  emp_sin 
                    AS sin,
                project_number
                    AS project_number2
        FROM    dbo.works_on ) AS w2 
    ON  w1.sin = w2.sin
    AND w1.project_number1 &lt; w2.project_number2;