Answer: A, B, C
The reducing power of HX rises from HCl to HI. HCl is not oxidised, HBr gives BrX2 and SOX2, and HI gives IX2 and HX2S. Fluoride cannot be oxidised by HX2SOX4 (E∘(FX2/FX−)=2.87 V), so NaF only gives HF. Options 1, 2 and 3 are correct.