I have showed that if $\alpha$ and $\beta$ are closed forms on a smooth manifold $M$, then $\alpha \wedge \beta$ also be closed. Further, if one of $\alpha$ or $\beta$ is exact, than $\alpha \wedge \beta$ is exact.
Than the question ask me to deduce that for $\alpha$ and $\beta$ are closed forms whose cohomology classes are $[\alpha]$ and $\beta$ respectively, there is a well-define cup product
$$\cup :H^p(M) \times H^q(M) \to H^{p+q}(M)$$
given by $[\alpha]\cup[\beta]=[\alpha \wedge \beta]$
Is the question ask me to show that $[\alpha]\cup[\beta]=[\alpha \wedge \beta]$ is well-defined only? So I pick $[\alpha]=[\alpha^\prime]$ and $[\beta]=[\beta^\prime]$ so that $\alpha = \alpha^\prime +\text{d}a$ and $\beta= \beta^\prime +\text{d}b$ respectively. And work my way through to show that
$$[\alpha]\cup[\beta]=[\alpha \wedge \beta]=[(\alpha^\prime +\text{d}a)\wedge(\beta^\prime +\text{d}b)]= \cdots = [\alpha^\prime]\cup[\beta^\prime]$$
Am I right?
Yes, this is what you're asked to show.