How can I calculate the amount of HF in aquoues solution of NH4F and an aquoues solution of NH4HF2?
WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts. 

Assuming:

(1) that concentrations can be used instead of activities (bad assumption...)
(2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations.
(3) Since HF is a relatively weak acid, little $\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$.

Using equilibrium:
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4.7$$

From 5 molar $\ce{NH4F}$, $\ce{[F-] = \pu{5 molar}}$

From 0.18 molar $\ce{NH4HF2}$, $\ce{HF_2^- = \pu{0.18 molar}}$

So plugging the number in:

$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}} = \dfrac{0.18}{\ce{[HF]}\times 5}=4.7$$

$$\ce{[HF] = 0.008 molar}$$

For the final concentrations $\ce{[HF] \ll [H2F-]}$, so the assumption that the dissociation of $\ce{H2F-}$ can be ignored is valid. The $\ce{HF}$ will dissociate and acidify the solution, but still $\ce{[HF] + [H+] \ll [H2F-]}$