I am stuck on a question, how does one show that for any ring $R$ if for all $a \in R$ there exists $n(a)>1$ natural such that $a^{n(a)}=a$ then $R$ is semi-primitive?
I suppose one should show that this implies that each element is not quasi-regular and thus the Jacobson radical is 0, but I am at a total loss on how to achieve this.
After thinking a bit about your original comment of proving only $0$ was quasiregular, I realize that leads to a very straightforward solution.
If $a$ is nonzero and right quasiregular, then $1-ar$ is a unit for every $r\in R$. But we always have that $a(1-ar)=0$ where $r=a^{n(a)-2}$. That prevents $1-ar$ from being a unit, and shows that the Jacobson radical is zero.
Original overkill:
Assuming you meant that $n(a) >1$ for all $a$, then it is obvious that the ring does not have any nonzero nilpotent elements. For, if $a$ were a nonzero nilpotent, then there is a power $a^i$, label it $b$, such that $b\neq 0$ and $b^2=0$. But since $n(b)\geq 2$, this says $b^{n(b)}=b=0$, a contradiction. Therefore the ring has no nonzero nilpotent elements (= a reduced ring.) The intersection of prime ideals is therefore the zero ideal.
Furthermore by a theorem of Jacobson, the ring is commutative.
Now let $P$ be a prime ideal. Then $R/P$ is a domain with the same property on its elements. In the quotient, whenever $0\neq a=a^{n(a)}$, we cancel a power of $a$ from both sides to get $1=aa^{n(a)-2}$ and see that $a$ is a unit. Therefore $R/P$ is a field, and $P$ is a maximal ideal.
Thus the intersection of maximal ideals is the zero ideal, and the ring is semiprimitive.