Derived functor of derived functor This is a dumb question.
Q1: Is there a reason to consider doubly/triply derived functors(i.e. derived functor of derived functors? Say given projective resolution and a right exact functor. One can obtain the left derived functor. Suppose to some degree left derived functor is left exact. Then I can further derive this left exact functor. Does this procedure have to stop at some time point? Of course, right derived functor of right derived functor might be trivial like Ext. 
Q2: Does the derived functors contain most of information on the complex except homotopy information? Any further derived functors do not yield extra information?
 A: Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $\mathrm{Tor}_1^A(M,\square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as  a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult. 
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$. 

Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
  $$
(R^n\mathrm{Tor}_1^A(M,\square ))(N)\cong\begin{cases} \mathrm{Tor}_1^A(M,N) \quad\text{when}\, n=0,\\tM\otimes_AN\qquad\,\,\text{when}\, n = 1,\\0\qquad\qquad\qquad \text{otherwise}.\end{cases}
$$

Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0\mathrm{Tor}_1^A(M,\square )(N)\cong \mathrm{Tor}_1^A(M,N) \quad\text{and}\quad (R^n\mathrm{Tor_1}^A(M,\square))(N)\cong 0\quad\text{for}\, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $\mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0\rightarrow N'\rightarrow N\rightarrow N''\rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0\rightarrow \mathrm{Tor}_1(M,N') \rightarrow \mathrm{Tor}_1(M,N)\rightarrow\mathrm{Tor}_1(M,N'')\rightarrow R^1\mathrm{Tor}_1(M,N')\rightarrow R^1\mathrm{Tor}_1(M,N)\rightarrow R^1\mathrm{Tor}_1(M,N'') \rightarrow 0.
$$
But thinking back to the LES of $\mathrm{Tor}$, we also have the LES:
$$
0\rightarrow \mathrm{Tor}_1(M,N') \rightarrow \mathrm{Tor}_1(M,N)\rightarrow\mathrm{Tor}_1(M,N'') \rightarrow M\otimes N'\rightarrow M\otimes N\rightarrow M\otimes N'' \rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1\mathrm{Tor}_1(M,N) \cong M\otimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $M\otimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $\Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition. 
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 \rightarrow N\rightarrow E_0 \rightarrow E_1\rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
\require{AMScd}
\begin{CD}
\mathrm{Tor}_1(M,E_0) @>>> \mathrm{Tor}_1(M,E_1) @>\delta >> R^1\mathrm{Tor}_1(M,N) @>>> R^1\mathrm{Tor}_1(M,E_0) \\
@| @| \\
\mathrm{Tor}_1(M,E_0) @>>> \mathrm{Tor}_1(M,E_1) @>\partial >> M\otimes N @>>> M\otimes E_0
\end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1\mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $M\otimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1\mathrm{Tor}_1(M,N)$ to $M\otimes N$. 
The general case follows from the torsion case by considering the SES
$$
0 \rightarrow tM \rightarrow M \rightarrow M/tM \rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $\mathrm{Ext}^1(M,\square )$ returns

Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
  $$
(L_n\mathrm{Ext}_A^1(M,\square )) (N) \cong \begin{cases} \mathrm{Ext}_A^1(M,N)\quad\text{when}\, n = 0,\\ 
\mathrm{Hom}_A(tM,N)\:\text{when}\, n = 1,\\
0\qquad\qquad\qquad\text{otherwise.}
\end{cases}
$$

Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $\mathrm{Tor}$ and $\mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive. 
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $\mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $t\cong \mathrm{Tor}_1^A(\mathrm{Frac}(A)/A ,\square )$, so that's something.
A: I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence 
$$0 \to A \to B \to C \to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) \to L^iF(A) \to L^iF(B) \to L^iF(C) \to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 \to A \to B$ one has that the map $L^{i-1}F(A) \to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact. 
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
A: The total derived functor is already exact, but in the relevant homotopical sense. 
In fact, you can say more; if $A \to B \to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' \to B' \to C'$ that is short exact in the ordinary sense.
 (by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object) 
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)

The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
