In which state does the bifluoride anion exist? Statement 1. The $\ce{HF2-}$ ions exist in solid state and also in
  liquid state but not in aqueous state.
  Statement 2. The magnitude of hydrogen bonding between $\ce{HF}$ is weaker than that between $\ce{HF}$ and $\ce{H2O}$  
  
  
  Both are correct and 2 explains 1  
  Both are correct but 2 does not explain 1  
  1 is correct and 2 false  
  2 is correct and 1 false   
  


The answer given is 1.

I know the second statement to be true. $$\Delta H (\ce{H-F...O})=40{-}45\text{ kJ/mol},\Delta H(\ce{H-F...F})=20{-}23\text{ kJ/mol}$$

But I am completely unsure about statement 1 and the relation of the two statements. I think the existence of the afforementioned ion in aqueous solution must not be absent, as I have read somewhere that this ion acts as the conjugate base of the molecule $\ce{H2F2}$ which dissociates into hydrofluoric acid.
Wikipedia page for for bifluoride anion. The question's answer appears to be true, with explanation and references.

The fluorohydrogenate group $\ce{-HF-}$ in anions such as bifluoride can assimilate a proton by recombination:

$$\ce{HF2- + H+ -&gt; H2F2 -&gt; 2 HF}$$

Because of this capture of a proton ($\ce{H+}$), bifluoride has basic character. Its conjugate acid is the reactive intermediate, &mu;-fluoro-fluorodihydrogen ($\ce{H2F2}$), which subsequently dissociates to become hydrogen fluoride. In solution, most bifluoride ions are dissociated.

$\ce{HF2- + 3H2O}$ is in a favored equilibrium with $\ce{2H3OF + HO-}$

This makes for an interesting oxygen coordination sphere re the other lone pair of electrons.  One presumes it is hydronium fluoride in which the fluoride is bonded to a proton with a delocalized charge.  

I've worked with aqueous ammonium hydrogen bifluoride as a silicate etchant. The solution has tremendous surface tension.  A drop of Fluorad surfactant was necessary to have things work.  $\ce{HF}$ will reach out and eat you.  $\ce{NH4HF2}$ is fairly benign.  No skin contact!  Don't work in Pyrex.