so I just do case that k is a even number right? and is it correct
3.not sure
$111...111=10^{k}+10^{k-1}+...+1=\frac{10^{k+1}-1}{9} \equiv (2\times5)^{k+1}-1\equiv 0(modn)$
$\because$n is odd and (n,5)=1$\therefore 10^{\phi{n}} \equiv 1(modn)$
$\therefore k=\phi{n} -1$