Distribution of palindromic numbers
In the interval $[100,999]$ there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\begin {cases} 9\cdot 10^{\frac {n-2}2} & n \text { even} \\ 9\cdot 10^{\frac {n-1}2} & n \text { odd} \end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways .
To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have four choices $(1-4)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $144$ numbers. There might be more, as I have avoided carrying.
See http://mathworld.wolfram.com/PalindromicNumber.html
The first bit of information you ask about is contained in the statement that $$\text{number of palindromes $\le 10^n$ }=\cases{2(10^{n/2}-1) & $n$ even \\ 11\cdot10^{(n-1)/2}-2 & $n$ odd }$$
(I'll have a think about the reverse-and-add algorithm if you're still interested in that.)