Congratulations, you've just discovered the definition of normal variety — precisely the one for which locally all rings are integral closures in their fields of fractions.
This definition leads to several inetersting geometric properties, e.g. the singular locus should have codimension at least 2. In particular, curves are normal iff they are nonsingular. See, for example, Rigorous Trivialities.
You second question is harder to understand, perhaps you could rework it a bit given this answer?
Here is a solution to the hint:
First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.
Next, the condition that $\mathfrak b_1 \cap \cdots \cap \mathfrak b_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b_1 \times \cdot \times
A/\mathfrak b_r$ (the product ot the natural quotient maps) is injective, while the
condition that $\mathfrak b_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b_i$, then $I \cap J \neq 0$ also.
Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I_j := \mathfrak b_1 \cap \cdots \cap \mathfrak b_{i-1} \cap \mathfrak c_j\cap \mathfrak b_{i+1}
\cap \cdots \cap \mathfrak b_r,$ for each $j =1,\ldots,s$.
Then $I_1\cap \cdots \cap I_s = 0$ (since it is contained in the intersection
of the $\mathfrak c_j$, which already vanishes).
Now we recall that $A$ embeds into the product of the $A/\mathfrak b_{i'}$.
Note that $I_j$ is contained in $\mathfrak b_{i'}$ for $i'\neq i$. Thus,
if we let $J_j$ denote the image of $I_j$ in $A/\mathfrak b_i$, then
we see that the image of $I_j$ under the embedding
$A \hookrightarrow A/\mathfrak b_1\times\cdots\times A/\mathfrak b_i \times \cdots \times A/\mathfrak b_r$
is equal to $0 \times \cdots \times J_j \times\cdots \times 0$.
Thus the intersection of the images of the $I_j$, which is the image
of the intersection of the $I_j$ (since we looking at images under an embedding)
is equal to $0\times \cdots \times (\bigcap J_j) \times \cdots \times 0.$
Thus, since the intersection of the $I_j$ is equal to $0$, we see
that $\bigcap J_j = 0.$ But $\mathfrak b_i$ is irreducible, and so
one of the $J_j = 0$. Equivalently, the corresponding $I_j = 0.$
This proves the hint.
(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)
Best Answer
Let B be the subring of K containing A. I am going to prove that B is the localization of A at a prime ideal $P\subset A$, which seems a reasonable interpretation of the statement that B is a local ring of A.
First B is a local ring [by Atiyah-MacDonald Prop 5.18 i)] ; let $M_B$ be its maximal ideal. Similarly let $M_A$ be the maximal ideal of A. Define $P=A\cap M_B$. Then of course $P\subset M_A$ is prime and if we localize A at P, I claim that we get B:
a) If $\pi \in A-P$ then $\pi \notin M_B$ and so $\pi$ is invertible in B. Hence for all $ a \in A$ we have $a/ \pi \in B$. This shows $A_P \subset B$.
b) Now we see that B dominates $A_P$: this means we have an inclusion of local rings $A_P \subset B$ and of their maximal ideals: $PA_P \subset M_B$. But $A_P$ is a valuation ring of K because A is (This is the preceding exercise 28 : these guys know where they are heading!) and valuations rings are maximal for the relation of domination (Exercise 27: ditto !). Hence we have the claimed equality $B=A_P$.