How does NaOH modify pyranodioxins?
The molecule in black consists of a cyclic ketal and a hemiacetal. The hemiacetal is known to be in equilibrium with its aldehyde form, which is the the molecule in blue, in aqueous solutions even without $\ce{NaOH}$. Recall the reaction of glucose (a cyclic hemiacetal) with dilute sodium hydroxide, which is known as Lobry de Bruyn-van Ekentein rearrangement (Ref.1).
Nonetheless, a hemiacetals can be decomposed to the relevant carbonyl and the alcohol under basic conditions as explained here (a reliable mechanism is given). In your case both carbonyl and the alcohol groups are within the same molecule (the molecule in blue):

All acetals and ketals are stable to alkaline conditions since they don't carry free hydroxyl group.

References:

Roy L. Whistler, J. N. BeMiller, &quot;Alkaline Degradation of Polysaccharides,&quot; Advances in Carbohydrate Chemistry 1958, 13, 289-329 (DOI: https://doi.org/10.1016/S0096-5332(08)60359-8).