Rules for Tetromino Tiling of a Rectangle Various Tetromino problems have been posted online. It has been claimed that
“If a rectangle with alternating black-white squares can not be filled with alternating black-white Tetrominos, then a solid-colored rectangle can not be filled with solid-colored Tetrominos”. Obviously, the number of squares must be a multiple of 4 for this even to be an issue. A 3x3 rectangle obviously can have alternating black-white squares and of course it can not be tiled with Tetrominos.
Is this true? Has it been proven? 
Given that all Tetrominos have 4 squares all of them can be comprised of two white and two black squares. However, the T Tetromino can not be covered with 2 white and 2 black squares without violating the alternation requirement. 
This gives rise to the following question.
“Does their exist any complete tiling (using Tetrominos) of any rectangle (of any size) that uses an odd number of T Tetrominos?”
If the answer to that question is NO, then such an answer would tend to support the first proposition.
 A: 
“If a rectangle with alternating black-white squares can not be filled with alternating black-white Tetrominos, then a solid-colored rectangle can not be filled with solid-colored Tetrominos”.
Is this true?

Yes, and it's easy to see why, provided I've understood the question correctly. If you can fill a (specific, given, $m\times n$) solid rectangle with solid-coloured tetrominoes, then do so. Then get out your black and white paints, and paint the tetromino-covered rectangle alternately black and white. This is a covering of the same ($m\times n$) black-white rectangle with black-white tetrominoes.

“Does their exist any complete tiling (using Tetrominos) of any rectangle (of any size) that uses an odd number of T Tetrominos?”

No, and I think you've guessed why: you should now be able to prove this easily by using the same method as the above.
(Suppose there did exist such a tiling. Assemble it, then paint it black and white. Now, T tetrominoes don't have the same number of black squares as white squares, and all the others do. This means your rectangle must not have the same number of black squares as white squares. This can only happen if the sides of your rectangle are both odd.)
