Concrete example of a dense Hamel basis of $C[a,b]$? See this old question: Does C[a,b] has a dense Hamel basis?
The answer provided by David C. Ullrich proves that every Banach space $X$ with $\dim X=\infty$ has a dense Hamel basis. 
The question now is: can I construct a concrete example of such dense basis? The basis used in Fourier expansion is clearly not dense in $C[a,b]$.
Actually, I am not limiting myself to  $C[a,b]$. Any Banach space can be fine.
 A: Of course there's no definitive answer to the question without a definition of "concrete". But, assuming it's like pornography, "can't give a definition but I know it when I see it": Surely there is no such basis. Obtaining any Hamel basis for an infinite-dimensional Banach space uses the axiom of choice, extremely non-concrete.
You say "The basis used in Fourier expansion is clearly not dense in $C[a,b]$." It seems likely this  indicates a major misunderstanding, perhaps explaining why you don't appreciate what's so difficult here:
Taking $[a,b]=[0,2\pi]:=I$ it appears you're talking  about $B=(e_n)_{n\in\Bbb Z}$, where $e_n(t)=e^{int}$. This is not just not dense in $C(I)$, it's not a Hamel basis for $C(I)$ in the first place.
It sounds like you've forgotten the definition. Because it's utterly and completely totally obvious that $B$ is not a Hamel basis for $C(I)$, just because there exists a continuous  function that is not a trigonometric polynomial.
In fact, although this is less  trivial, $B$ is not even a Schauder basis for  $C(I)$; if  it were that would say the Fourier series for any continuous function converges uniformly, which is not so. $B$ is a Schauder basis for $L^2(I)$, although again not a Hamel  basis.
It  seems clear to me that no infinite-dimensional Banach space has a "concrete" Hamel basis. As mentioned, we can't prove that until we define "concrete", but the following is well known and easy:


Fact If $E$ is an infinite-dimensional Banach space then any Hamel basis is uncountable.


Proof: In fact  any spanning set must be uncountable. Say $b_1,b_2,\dots\in E$. Let $V_n$ be the span of $b_1,\dots, b_n$. Then $V_n$ is closed  and has empty  interior,  so the Baire Category Theorem implies $$\bigcup_n V_n\ne E.$$
