Calculation of Ext Let $A$ be an abelian group.
I know that $Ext_\mathbb{Z}^1(\mathbb{Z}/p,A)=A/pA$.
Are there any similar formula about $Ext_\mathbb{Z}^1(A,\mathbb{Z}/p)$?
I know that $Ext_R^n(A,B)\neq Ext_R^n(B,A)$ generally.
 A: Form the injective resolution $$0 \to \dfrac{1}{p}\mathbb Z/\mathbb Z \to
\mathbb Q/\mathbb Z \buildrel \cdot p \over \to \mathbb Q/\mathbb Z \to 0.$$
Applying $\mathrm{Hom}(A,\text{--})$ to this, we find that 
$$\mathrm{Ext}^1(A, \mathbb Z/p) = 
\mathrm{Ext}^1(A,\dfrac{1}{p}\mathbb Z/\mathbb Z)
= {\mathrm Hom}(A,\mathbb Q/\mathbb Z)/p.$$
Note that $\mathrm{Hom}(A,\mathbb Q/\mathbb Z)$ is (speaking somewhat informally; see my comment addressed to Rasmus below) a version of the
Pontrajagin dual of $A$.  (In particular, if $A$ is torsion, then it
is precisely the Pontrjagin dual.)  In fact, we could replace $\mathbb Q/\mathbb Z$
by the circle group $S^1$ in the above injective resolution, and then 
we would find that
$$\mathrm{Ext}^1(A,\mathbb Z/p) = \hat{A}/p\hat{A},$$
where $\hat{A} = \mathrm{Hom}(A,S^1)$ is the Pontrjagin dual of the abelian group $A$ (with its discrete topology).
[Note: I began writing this answer some hours ago, but then got distracted.
It is essentially the same as that of Rasmus, although, rather than using
either $\mathbb Q/\mathbb Z$ or $S^1$, he uses $\mathbb Z(p^{\infty})$, which is
the $p$-power torsion part of $\mathbb Q/\mathbb Z$.  (If you like, you can write it as $\mathbb Q_p/\mathbb Z_p$.)  These different descriptions reflect the flexibility available in choosing an injective resolution of $\mathbb Z/p$.]
A: We need an injective resolution for $\mathbb Z/p$ in the second variable. An Abelian group is injective if and only if it is divisible.
So, the first step will be to embed $\mathbb Z/p$ into a divisible Abelian group. For this, we can use the Prüfer $p$-group $\mathbb Z(p^\infty)$.
The cokernel of this embedding is again the Prüfer $p$-group, so that we get the injective resolution $\mathbb Z(p^\infty)\to\mathbb Z(p^\infty)$ given by multiplication with $p$.
Hence we get
$$
Ext_\mathbb{Z}^1(A,\mathbb{Z}/p)=\frac{Hom_\mathbb{Z}(A,\mathbb{Z}(p^\infty))}{p\cdot Hom_\mathbb{Z}(A,\mathbb{Z}(p^\infty))}.
$$
I don't think that it is easy to understand homomorphisms into $\mathbb{Z}(p^\infty)$ because it is a direct/injective limit. (Edit: But please read Matt E's comment below.)
A: I would just add that $\operatorname{Ext}(A,G)$ is fully computable when $A$ and $G$ are finitely generated abelian groups.
You can find the proofs in any decent book on homological algebra:
1) If $A$ is free abelian, then $\operatorname{Ext}(A,G)=0$
2) If $G$ is divisible $\operatorname{Ext}(A,G)=0$
3) $\operatorname{Ext}(A,\prod G_j)=\prod \operatorname{Ext}(A,G_j)$
4) $\operatorname{Ext}(\sum A_j,G)=\prod \operatorname{Ext}(A_j,G)$
5) $\operatorname{Ext}(\mathbb{Z}/m\mathbb{Z},G)=G/mG$ (as noted)
