Why is $((aa)^*bb(aa)^*bb(aa)^*)^*$ of star-height 1
(This is an attempt to an answer, I hope the details are right.)
Your language consists of all strings in $(aa+bb)^*$ with an even number of $bb$.
We are allowed to use complementation, so we start by looking at the complement of the language. I think we can split the complement into two (overlapping) parts

strings not of the form $(aa+bb)^*$, that language has starheight one.
strings of the form $w_0\cdot bb\cdot w_1\cdot bb\cdot w_2 \dots bb\cdot w_n$, where (1) the number of $bb$ is odd, and (2) the $w_i$ do not contain any $bb$ (but I will have to be more precise on that)

Now we will try to find an expression for the language $W$ of the $w_i$ without using star. That means we can use the star to count the odd number of $bb$, as in $(W\cdot bb\cdot W\cdot bb)^*\cdot W\cdot bb\cdot W$.
This is the tricky part. Strings from $W$ are either

the empty string $\varepsilon$
just one single $a$
of the form $axa$, where $a$ does not contain $bb$. Not &quot;containing $bb$&quot; is the complement of $(a+b)^*bb(a+b)^*$, and as you know $(a+b)^*$ is again star-free.

The strings we now have specified can have longer stretches of $b$'s, but whenever that happens the $b$'s come in pairs.