Is $Z_{p^2}$ a $Z_p$ module? $Z_{p^2}$ is a free $Z_{p^2}$-module. how can i see if it's a $Z_p$-module?
Is $Z_p \bigoplus Z_p$ a $Z_p$-module? a $Z_{p^2}$-module? 
In general, is there any general methods for these type of questions? thank you.
 A: First, advice: don't use $Z_p$ to mean the quotient $\mathbb Z/p\mathbb Z$ or $\mathbb Z/p$, since $\mathbb Z_p$ is very often the $p$-adic integers, not $\mathbb Z/p$. But, yes, many people save the vinculum...
Also, although textbooks and classes ask questions like this, in a nearly null context, in a vacuum, such things are rarely a genuine issue. That is, we don't have to "make up" an action of a ring on an abelian group, there is one naturally present. Perhaps we have to "see" the natural thing, and sometimes the natural thing is not as elementary as one's expectations suggest.
That is, I'd be interested in questions that mean "is there a natural action of this, on that", but quite disinterested in "is there any action of this on that". The answer to the latter might be difficult, but boring, and/or lead to typically-pointless questions about whether a module over a ring should be "unital" (meaning that $1\cdot m=m$ for all $m$ in the module). 
That is, to exaggerate only slightly, I'd answer such questions with the question "Why do you ask?" ... as a mild parody of asking for the context.
In real life, usually modules over rings arise in a fashion that makes it obvious that they are such. It's not an issue of demonstrating this... except in the extremely artificial environments of certain classrooms. Nevertheless, yes, on some rare occasions, it is a significant insight to realize that a "thing" admits an action of a certain ring on it, thus giving it more structure than one had previously understood.
After that "real-life" advice, probably you were supposed to do your exercise by noting that a unital module over $\mathbb Z/p$ has the property that $p\cdot m=0\cdot m=0$. Since this does not hold in $\mathbb Z/p^2$, it cannot be a unital $\mathbb Z/p$-module.
The better part of the issue is, as noted in comments above, that a direct sum of two modules over a ring is easily, for good reasons, for formal reasons, a module over the same ring.
The answer to the needlessly general (not-necessarily-unital) version of the question is indeed answered by @BorisNovikov, and does have considerable interest, but perhaps is a bit of a tautology, and only intelligible at a slightly later point in the questioner's trajectory.
A: "In general, is there any general methods for these type of questions?" --
This is so called "change of rings" [Cartan, Eilenberg, Homological Algebra, Sec.2.6]: if $A$ is a $R$-module and $f:S\to R$ is a morphism then $A$ turns into a $S$-module by $s\cdot a=f(s)a$ for $a\in A, s\in S$.
So if you want to transform a $Z_{p^2}$-module into  a $Z_{p}$-module, you have to chose some morphism $Z_{p}\to Z_{p^2}$ (of course it is not unique).
A: Hint for your second question: is $\mathbb{R}^2$ an $\mathbb{R}$-vector space? If yes, then it has to be an $\mathbb{R}$-module as well. What is the multiplication map $\mathbb{R} \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$? This will help you showing that $\mathbb{Z}_p^2 = \mathbb{Z}_p \oplus \mathbb{Z}_p$ is a $\mathbb{Z}_p$ module and in fact a vector space if $p$ is prime.
To see if $\mathbb{Z}_p^2$ is a $\mathbb{Z}_{p^2}$-module (not a vector space this time), my hint is to check the existence of a canonical map $\mathbb{Z}_{p^2} \rightarrow \mathbb{Z}_p$. To see that, note that $\mathbb{Z}_p \cong \mathbb{Z}/p \mathbb{Z}$ and consider the canonical map $\mathbb{Z} \rightarrow \mathbb{Z}/p\mathbb{Z}$. What can you say about the kernel of this map and $p^2 \mathbb{Z}$?
In general, if you have an abelian group $M$ and a ring $A$, and you want to show that $M$ is an $A$-module, then you need to show the existence of a ring homomorphism from $A$ to the group endomorphisms of $M$.
