This is a somewhat stupid case for Miller-Rabin because you should assume that n is not a square. Anyway. Let $n=25$ and $a=2$ the smallest possible candidate, then $n-1=d\times 2^s = 3\times 2^3.$ Now compute $$x_0 \equiv a^d \equiv 2^3 \equiv 8 \pmod{25}$$ $$x_1 \equiv x_0^2 \equiv 14 \pmod{25}$$ $$x_2 \equiv x_1^2 \equiv 21\pmod{25}$$ Thus $a^d\
ot \equiv 1 \pmod {25}$ and $x_r \equiv a^{2^r d}\
ot \equiv -1 \pmod {25}$ for $0\le r \le s-1.\,$ Therefore $a=2$ is a witness for the compositeness of $25$.
Regarding the second part: Obviously, for general $n\,$ you **cannot** always find a _Miller-Rabin witness for the compositeness of n!_ (otherwise there would be no primes!)