You do still require that x>-2 though. That's indisputable, given that part of the equation involves log(3x+6) and that log is only defined on positive numbers.
Besides, I think the last part of b(i) needs the fact that x>-2. Knowing that x>-2 and x<-1.995 gives you a nice little bound on what alpha can be to 3sf.
Part bii wants you to apply a very basic principle that you can do with your calculator. Start off with some value of x, any you like so long as it's not too close to alpha. Type that into your calculator and press the equals button. Now that will be stored in the memory of the calculator that you can recall as "ANS". Are you familiar with this concept? Now, type into your calculator 0.25log(3ANS + 6). This will give you the first "iterate" and that number will now be stored as ANS. Again, type in 0.25log(3ANS + 6) to get the second iterate. As you keep applying this process you'll home in on the true root of the equation x=0.25log(3x+6).
Just so you have something tangible (I suppose I use the word "tangible" loosely
) to aim for, I started off with x=1, and I very rapidly homed in on the root 0.205139...