For 3rd one
$S\rightarrow aaS_{1}b|aS_{1}ab|aS_{1}ba|abS_{1}a|bS_{1}aa|baS_{1}a$|SS
$S_{1}\rightarrow a$
It is very easy and correct grammer, if u see all it's combinations.
Their explanation is also correct. I just add some more steps for easy explanation.
Suppose , a grammar follows 3rd rule , what it includes?
$(aaab,aaaaabb,bbaaaaa, ...................)$
So, we can do grammar like this,$S\rightarrow aaSb$, where we can get any $a^{2n}b^{n}$ grammar. But we need $a^{2n+1}b^{n}$. For that odd 1, we add it at last in CFL.