*Answer*: (a) Miku; (b) Juku.

The positional representation with radix $a$ consisting of $n$ ones denotes the sum $a^{n-1} + a^{n-2} + \dots + a + 1$ which equals $\frac{a^n-1}{a-1}$.

a. Let $a \equiv 1 \pmod{p}$. If $p > 2$ or $a \equiv 1 \pmod{4}$ then, by the lifting-the-exponent lemma, the exponent of the prime $p$ in the canonical representation of $\frac{a^n-1}{a-1}$ equals that in the canonical representation of $n$. Thus to win, Miku may choose any number $n$ that is divisible by $p$. If $p = 2$ and $a \equiv -1 \pmod{4}$ then the exponent of the prime 2 in the canonical representation of $a-1$ is 1, while that in the canonical representation of $a^2-1$ is larger as $a^2 \equiv 1 \pmod{4}$. Thus to win, Miku may choose $n = 2$.

Let now $a \not\equiv 1 \pmod{p}$. By Fermat's little theorem, $a^{p-1} \equiv 1 \pmod{p}$. Hence $a^n \equiv 1 \pmod{p}$ whenever $n$ is a multiple of $p-1$. Then the numerator of the fraction $\frac{a^n-1}{a-1}$ is divisible by $p$ while the denominator is not, whence the value of the fraction is divisible by $p$. Consequently, Miku can win by choosing any multiple of $p-1$ greater than 1 as $n$.

b. We show that Juku wins by choosing $n = 2p-1$. Let $a$ be the number chosen by Miku.

Let $a \equiv 1 \pmod{p}$. If $p > 2$ or $a \equiv 1 \pmod{4}$ then, by the lifting-the-exponent lemma, the exponent of $p$ in the canonical representation of $\frac{a^n-1}{a-1}$ equals that in the canonical representation of $n$. As $p$ does not divide $2p-1$, it does not divide $\frac{a^n-1}{a-1}$ either. If $p = 2$ and $a \equiv -1 \pmod{4}$ then $a^n \equiv -1 \pmod{4}$, whence 2 does not divide $a^n-1$. Therefore 2 does not divide $\frac{a^n-1}{a-1}$.

Let now $a \not\equiv 1 \pmod{p}$. By Fermat's little theorem, $a^{p-1} \equiv 1 \pmod{p}$ and $a^p \equiv a \pmod{p}$, giving $a^n = a^{2p-1} \equiv a \pmod{p}$. Thus $a^n \not\equiv 1 \pmod{p}$. As $p$ does not divide the numerator of the fraction $\frac{a^n-1}{a-1}$, it cannot divide the value of the fraction either.