Hardness of noisy Boolean functions
for Question 1, the answer is Yes, and can be shown as follows.  (I will also be implicitly sketching an affirmative answer to Q4, since the argument is uniform and will treat all input lengths at once.)  

Fix any NP-complete language $L$, and a family of good binary error-correcting codes (with rate 1/4 and correcting from a .1 fraction of errors, say).  Let $Enc_n: \{0, 1\}^n \rightarrow \{0, 1\}^{4n}$ be the encoding function for length $n$; we use some such code where $Enc = \{Enc_n\}$ is computable by a uniform polynomial-time algorithm.

Define $L'$ as the set of strings $z$ that are within distance at most $.05 |z|$ from a codeword $y \in Enc(L)$ encoding some element of $L$.  Note that $L'$ is in NP, as you can guess and check the nearby codeword, the decoded word, and the NP certificate for the decoded word's membership in $L$.  

Then the claim is that any "approximation" of $L'$ in your sense is NP-hard for $\varepsilon = .01$.  For if we consider a valid codeword $y = Enc(x)$ of some length $4n$, then with probability $1 - o(1)$ over a random $\varepsilon$-perturbed version $y'$ of $y$, it will disagree with $y$ in at most a .05 fraction of coordinates, and will therefore disagree with any other codeword of $Enc_n$ in a more than $.05$ fraction of coords.  For such $y'$ we have $y' \in L'$ iff $x \in L$.  So if $h$ is any approximation to the $\varepsilon$-smoothed version of $L'$ in your sense, then we must have $h(y) = L(x)$.  As $Enc$ is efficiently computable, this gives us a way to efficiently reduce membership questions for $L$ to ones for $h$.  So $h$ is NP-hard.

Two notes:  

(1) error-correcting encodings of NP instances have been used before in several papers, notably
D. Sivakumar: On Membership Comparable Sets. J. Comput. Syst. Sci. 59(2): 270-280 (1999).

(2) the argument above of course says nothing about the average-case complexity of any NP problem, since error-correction is being applied on an instance-by-instance basis.