dont really understand what is going on from the 5th line onwards...

This will be hard to explain online because there are lots of things going on in factorisation. It's much easier if this could be done on an interactive whiteboard.

In the fifth line, I merely re-positioned the "1/2" to the front of the numerator.

In the sixth line, I introduced the terms in purple (1/2 and 2) because I wish to factorise a "1/2" out later on. In doing so, I am trying to remove all the "fraction in the numerator or fraction in the denominator".

Also, in the sixth line, I split x^0.5 into x^-0.5 times x^1 because I wish to factorise out x^-0.5 as well.

Remember that common terms will be factorised out in the same way you would have already done in Sec 2, but here it is more difficult because there are powers and other stuff to consider. This is the reason why it's extremely difficult for me to explain it by text where facilities are limited.

Now I have prepared the necessaries for the factorisation step.

In the seventh line, I factored out the common "1/2" and the common "x^-0.5" in the numerator. This leaved behind a "3 + x" for the term before the minus sign and a "2x" after the minus sign in the denominator. I kept the denominator intact.

In the eighth and ninth lines, I simply polished up the numerator to make it look nicer.

Finally, in the tenth line, the "1/2" in the numerator goes down to the denominator as "2" (think why), while the x^-0.5 can be written in positive index form as 1 / x^0.5 in the numerator and thus this goes down to the denominator as x^0.5 in the denominator.

why cant we just factorise x^-1/2 out and cancel (3+x) tho... thats what i did but i got the answer wrong

You mean for Q2a (my other working)?

Remember that we need to divide both the terms in the numerator by (3 + x) if we wish to cancel the (3 + x).

I prefer to leave it undivided as our answers are best left behind in a single fraction.

yep, just realised i commented on the wrong answer

we cant just do the normal cancellation of common terms? why do we need to divide the other terms at the numerator with (3+x) if we wish to cancel tho

Imagine you have 40 sweets and 24 balloons. You wish to divide these equally amongst 8 students. How many sweets and balloons will each student get?

Well, we can represent this as

(40 sweets + 24 balloons)

divided by 8.

We need to divide the 8 on both the 40 sweets and the 24 balloons.

Similar idea applies for the expression in the question.

how about factorising (x^1/2) to change the power of x^1/2 into x^1, i did that and got another answer instead...

You must rewrite x^1/2 as the product of x^-1/2 and x^1, because 1/2 = -1/2 + 1

ohh, btw if we want to bring the half down we have to make sure the half is factorised out from all the terms in the numerator? i did it without introducing 1/2 and 2 to x^-1/2...

You need to ensure that expressions are factorised out first before you can do direct division.