How to show that $\operatorname{Hom}_A(A/(b), K/A)$ is isomorphic to $A/(b)$? 
Let $A$ be a integral domain and let $K$ be its field of fractions.
Show that for any $b \in A$,
$\operatorname{Hom}_A(A/(b), K/A)$ is isomorphic to $A/(b)$.

So I have a few questions.
1. I don't quite understand the notation $K/A$. Can someone explain it to me?
2. How do I begin the first step, that is to construct a homomorphism ?
 A: The following general fact is well known: $\operatorname{Hom}_R(R/I,M) = \{m \in M ~|~ Im=0\}$.
Let $b \neq 0$.
In our case, this yields $\operatorname{Hom}_A(A/(b),K/A) = \{x \in K/A ~|~ bx=0\}$.
$bx=0 \in K/A$ means $bx \in A$, i.e. $x = \frac{a}{b} \in \langle \frac{1}{b} \rangle$ for some $a \in A$. Thus we have shown the desired
$$\operatorname{Hom}_A(A/(b),K/A) = \{x \in K/A | bx=0\} = \langle \frac{1}{b} \rangle \cong A/(b).$$

For the sake of completion: For $b=0$, we have 
$$\operatorname{Hom}_A(A/(b),K/A) = \operatorname{Hom}_A(A/,K/A) = K/A \not\cong A = A/(b),$$
i.e. the statement is true for $b \neq 0$, but false for $b=0$.
A: You have to assume $b\ne0$, because in this case the result is false.
Given $a\in A$, you can define $\tilde{a}\colon A\to K/A$ by
$$
\tilde{a}(x)=\frac{ax}{b}+A
$$
(when $b\ne0$). It is clear that $(b)\subseteq\ker\tilde{a}$, so $\tilde{a}$ defines $\hat{a}\colon A/(b)\to K/A$ by
$$
\hat{a}(x+(b))=\frac{ax}{b}+A
$$
It's easy to verify that $a\mapsto\hat{a}$ is a homomorphism
$$
\varphi\colon A\to \operatorname{Hom}_A(A/(b),K/A)
$$
If $a\in (b)$, then $\tilde{a}$ is the zero map, so also $\hat{a}=0$. Conversely, if $\hat{a}=0$, we have
$$
\hat{a}(1+(b))=\frac{a}{b}+A=A
$$
that is, $a\in(b)$. Thus $\varphi$ induces a monomorphism
$$
\tilde{\varphi}\colon A/(b)\to\operatorname{Hom}_A(A/(b),K/A)
$$
and it remains to show that $\varphi$ is surjective. This will follow when we show that any homomorphism $f\colon A\to K/A$ that vanishes on $(b)$ is of the form $\tilde{a}$, for some $a$. Let
$$
f(1)=\frac{t}{u}+A
$$
so $f(x)=\frac{tx}{u}+A$. By assumption $f$ vanishes on $b$, so
$$
\frac{tb}{u}\in A
$$
and we are done by choosing $a=tb/u$.
