1. Riboflavin
Walsh & Tang state about Leukoflavin that its "three ring conjugated double bond system [is] interrupted" thus "leuko" (Greek for “white” or colourless; because no visible light absorption any more). 
Where in the structure is the interruption? Don’t we still have the same nitrogen hybridizations as in Riboflavin, thus suggesting conjugation? Or is it that previous diimine (N = sp2) is now a N-sp3 in Leukoflavin?

Indeed, in FADH2, the NH1-C10a & C4a-NH5 bond nitrogens are explicitly sp3 so not conjugated (what geometry would you expect at Nitrogen? flat or bent? you will see a picture in Rottlant; note that “gas phase” geometry means the geometry that should be seen when there is no restriction by enzyme binding pocket shape). 
	Compare this to the N1=C10a-C4a=N5 bridge (conjugated) of the oxidised state (numbering in Walsh Fig 4.1), which keeps both nitrogens explicitly sp2. 

Yes, nitrogens can be tricky because we sometimes "bring a ring N lone pair into conjugation” even when the nitrogen only has single bonds (like in pyrrole) - but we only do it if this would create an aromatic system (4n+2 π electrons). In this case, it would not. So, no shared lone pair, no conjugation, leuco system.

At least this is the answer in our simple atomic orbitals picture, which is what we use in courses. 

If you want a better analysis than just atomic orbitals, the molecular orbitals picture will be more complicated but it is much more satisfying. You find some highlighted passages here in Rotllant (along with molecular structures, and absorption spectra) and they should be helpful!

2. Thiamin
Transketolase: How is it decided that in our pentose product (D-Xylulose-5P), hydroxy group at C3 is on the left side (L stereochemistry in Fischer) after addition of acyl anion to D-Glyceral-3P's aldehyde group? 
Is there a possibility of getting a C3 with a D configured hydroxy group?

I believe that this is for the same reason as in all other enzymes I know: the coenzyme sets the chemistry, but the protein geometry and interactions around the active site preorient the substrates relative to each other, so that one stereochemical outcome dominates. 

If you would mix TPP and hydroxyketone and aldehyde in free solution with no enzyme, I am sure you would get both L and D exactly like you say. And the D product would probably be useless for the next steps in biology.