For (a), your resolution is not right, because $(\prod_{p\in S^1}i_p(\mathbb{Q}))/\mathbb{Z}$ is not $\prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$. Or in other words, the kernel of $\prod_{p\in S^1}i_p(\mathbb{Q})\rightarrow \prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$ is not $\mathbb{Z}$ (this is $\prod_{p\in S^1}i_p(\mathbb{Z})$).
For (b), you cannot mimick example 4.0.4 to compute the Cech cohomology. In fact, the higher Cech cohomology of this sheaf vanishes. Let $U,V$ your cover (which is good), and consider the Cech complex
$$ R(U)\times R(V)\rightarrow R(U\cap V)$$
given by $(f,g)\mapsto f_{|U\cap V}-g_{|U\cap V}$.
The kernel of this map are couple of functions $(f,g)$ that agree on the intersection, and hence patch together to form a unique function on $S^1$. So $\overset{\vee}{H^0}(\{U,V\},R)=R(S^1)$ as expected.
But I claim that the map is onto, so that $\overset{\vee}{H^1}(\{U,V\},R)=0$. Indeed, let $f\in R(U\cap V)$, also let $u,v$ be two functions on $S^1$ such that $u$ has (compact) support in $U$, $v$ has (compact) support in $V$ and $u+v=1$. In particular the function $u$ is zero in a neighborhood of $S^1\setminus V$ and $v$ is zero in a neighborhood of $S^1\setminus U$. The function $uf$ can then be extended to $U$ because it is zero in a neighborhood of $U\setminus U\cap V$. Similarily, the function $vf$ can be extended to $V$. And $(uf,-vf)\mapsto uf+vf=f$ so that the map is onto.
Here $u,v$ are called a partition of unity. A sheaf like $R$ with partitions of unity is called a fine sheaf. This argument (of a similar one with more than two open sets in the cover) shows that the higher Cech cohomology groups of a fine sheaf over a paracompact space vanish. On a paracompact space, Cech and derived functor cohomology agree so the cohomology of a fine sheaf is trivial.
The difference with the example 4.0.4 is that there is no partition of unity in a constant sheaf.
Best Answer
I suspect (but am not sure) that your article may be referring to hypercohomology. The point of hypercohomology is that, given a functor $F: \mathcal{A} \to \mathcal{B}$ (say, left-exact, like the global section functor; let's also assume $\mathcal{A}$ has enough injectives), one can define the so-called "hyper-derived functors" $\mathbf{R}^i F$, each of which is a functor from complexes on $\mathcal{A}$ to $\mathcal{B}$. A short exact sequence of complexes leads to a long exact sequence of hypercohomology, just as with the ordinary derived functors.
The more modern way to think of hypercohomology is to use the derived category. The point is then that a functor $F: \mathcal{A} \to \mathcal{B}$ induces a total derived functor functor on the bounded-below derived categories $\mathbf{D}^+(\mathcal{A}) \to \mathbf{D}^+(\mathcal{B})$ (you can think of the derived category as localizing the category of chain complexes with respect to quasi-isomorphisms, though it's better to go first through the homotopy category). Then the hypercohomology functors are just defined by taking the $i$th cohomology of the total derived functor.
To compute this, you start with a bounded-below complex $K^\bullet$, find a quasi-isomorphism $K^\bullet \to I^\bullet$ where $I^\bullet $ consists of $F$-acyclic (say, injective) objects, and take $F(I^\bullet)$ as the output of the derived functor.
I could say more if you clarify that this is in fact what you are looking for!