Prove reciprocal polynomial is not primitive(field). Let $f$ be a polynomial with a non-zero constant term and define its reciprocal polynomial $f^*=x^nf(\frac{1}{x})$.
(Here $\deg {f}=n$. I see several other names like "reverse/inverse polynomial", and in the following case "self-inverse".)
We say a polynomial $f$ is primitive over a field $F_p$(p is prime) iff $f$ is irreducible and has its degree $n$ with $k=p^n-1$the lowest one satisfies  $f|x^k-1$. Please prove the following:  

$f=f^*$and $\deg f>2$ $\Rightarrow$ $f$ is not primitive. 

I don't have any specific idea. If some properties or definition of the primitive polynomial are used, you can just state it. And hints will also be apprciated.
Thanks in advance.  
Edit: assuming $\deg f>2$
Edit2: the definition of "primitive"
 A: Original Answer: The claim is false.
Take $p=2$, the polynomial $x^2+x+1$ is irreducible over $\mathbb{F}_2$ (hence primitive) and self-reverse.

Addendum: With the constraint $\deg f>2$ the result is still false.  The self-reverse polynomial $x^4+x^3+x^2+x+1$ is irreducible degree 4 over $\mathbb{F}_2$ so is primitive.
There seems to be something wrong with your definition of primitive.  Over finite fields, an irreducible polynomial having a root in a finite extension implies the polynomial splits completely in that extension, we have $f\mid x^{p^n-1}-1$ automatically if $f\in\mathbb{F}_p[x]$ irreducible degree $n$.

Addendum 2: Now with the new definition of primitive it is easy.  Every self-reverse irreducible polynomial $f$ of degree $2n$ over $\mathbb{F}_p$ is actually a factor of $$
x^{p^n+1}-1\in\mathbb{F}_p[x]
$$
(in fact also works for $q$) because if $\lambda$ is a root of $f$, then $\lambda^p, \lambda^{p^2}, \dots, \lambda^{p^{2n-1}}$ are the other roots of $f$.  Because this set is invariant under inversion, $\lambda^{-1}=\lambda^{p^j}$ for some $j\in\{0,1,\dots,2n-1\}$.  Do that again, $$(\lambda^{p^j})^{-1}=(\lambda^{-1})^{p^j}=(\lambda^{p^j})^{p^j}=\lambda^{p^{2j}},$$ so we must have $2j=2n$ and hence $f\mid x^{p^n+1}-1$.
