You do know that you posted it upside down, don't you? It's also in Swedish which I, at least, cannot read. Surely, it wouldn't have been that hard to just type in the problem yourself.
I assume that you to describe, or write in a more simplified way, the sets
a) $\displaystyle \{x\in R: (x-1)(x-5)\le 0\}\cap\{x\in R: (x- 3)(x- 1)> 0\}\cap \{x\in R: (x- 2)(x- 6)\le 0\}$
b) $\displaystyle \{z\in C: Re((z-1)(\overline{z}+ i)= 1/2\}$
The first is simply asking for all numbers that satisfy all three of the inequalities. Do you know how to find numbers satisfying the inequalities separately? For example (x- 1)(x-5)= 0 when x= 1 or x= 5. Since the product is a continuous function of x, only those numbers can separate "<" and ">". In particular, if x= 2 (between 1 and 5), (2- 1)(2- 5)= (1)(-3)= -3< 0 so does satisfy the first inequality. You can check a number less than 1, say 0, and a number larger than 5, say 6, to see that numbers less than 1 or larger than 5 do NOT satisfy it. That is, the first set is simply the interval $\displaystyle 1\le x\le 5$ of all number between 1 and 5, including the endpoints. Find the solution sets of the other two inequalities in the say way and see what numbers, if any, satisfy all three.
(b) is harder. In fact, I am surprised you would have trouble with (a) if you are in a course where you are expected to be able to do (b). I would write z= x+ iy so that $\displaystyle \overline{z}= x- iy$. Then $\displaystyle z- 1= (x- 1)+ iy$ and $\displaystyle \overline{z}+i= x+ i(1- y)$ and then $\displaystyle (z-1)(\overline{z}+i)= [(x- 1)+iy][x+ i(1- y)]= (x(x-1)- y(1- y))+ i(xy+ (x-1)(1- y))= (x^2+ y^2- x- y)+ (x+ y- 1)i$
Find the real part of that and set it equal to 1/2. What is the graph of that in the complex plane?
No i did not know it was upside down because i did uppload through My phone:P ... I did translate in My text:S the thing is i did start same with a as u did but u dont have to say like that.. I did never read this in My book actually so... Im actually searching for this in My book but cant find.. Ima see what i can progress with