[StrongAcid22]

Is the reason for the corrosive nature of hydrofluoric acid to glass, deal with a property of the acid or the nature of the F^- ion?

If the later is correct, would not solutions of sodium and potassium fluoride need to be stored in Teflon containers?

Also, in a solution of sodium fluoride in which HF is a product of the hydrolysis, I am assuming that the little HF that dissolves reacts with the OH^- that was produced by the hydrolysis rather than the water. Is that correct?

[constant thinker]

Have you checked the MSDS to find out if it's incompatible with glass?
http://www.solvaychemicals.us/static/wma/pdf/5/1/5/7/NAF.pdf
http://www.solvaychemicals.us/static/wma/pdf/1/6/3/8/9/KF_CN_EN.pdf

Section 7 is handling and storage. Section 10 is Stability and Reactivity. One of those sections should help you answer your second question.

For your third question write out the reaction and equilibrium equation. Note that HF is a weak acid with a pK_a ~ 3. For comparison sulfuric acid has a pK_a1 ~ -3 and pK_a2 ~ 2. How would you go about carrying out this hydrolysis?

[StrongAcid22]

The hydrolysis is:

F^- + H₂O HF + OH^-

My question was basically if in the solution of sodium fluoride, if it was possible for the reverse reaction to react with the water rather than hydroxide ions from the hydrolysis.

[constant thinker]

The reaction is in equilibrium (like almost every other process). It's easier to write the reaction like this:
HF    H⁺ + F^-

Use the equilibrium equation to find out how much HF will form:
K_a=([H⁺][F^-])/[HF]

Assuming you're adding a known amount of NaF you can calculate how much HF would form, and what the pH would be.
Hint: Use an I.C.E. (Initial, Change, Equilibrium) chart. Is there a significant amount formed at an initial pH of 7 (assuming you're using DI water)? To keep the math simple it's ok to use 1M as your initial NaF concentration since this is just a theoretical treatment of the problem.

Your reaction is correct except you didn't write it as an equilibrium. It's important to remember that this an equilibrium reaction.
The equilibrium equation for the reaction you wrote would be:
K_w/K_a=K_b=([HF][OH^-])/[F^-]

Note: pK_x=-log(K_x) therefore K=10^-pK_x

[Borek]

You need to take into account water autoionization, otherwise you have only 10^-7M H⁺ - and it may seem [HF] can't go past 10^-7M, which is not true (in 1M solution of NaF concentration of HF is almost 100 times higher, close to 10^-5M)