Would this be sufficient for the first one:
Assume N_{n}N_{m}, then (t)N_{n}=N_{m}, t≥1. For all m,n≥1, there exists q,r such that m=qn + r, 0≤r<n.
Next assume m does not divide n, therefore 1≤r<n.
Thus N_{m}=N_{qn+r}=10^{r}(s)N_{n}+N_{r}
Then, (t)N_{n}=N_{qn+r}=10^{r}(s)N_{n}+N_{r}
Thus, N_{r}=N_{n}(ts10^{r})=N_{n}(d), d≥1, which implies r≥n, which contradicts our hypothesis of r<n.
Thus N_{n}N_{m} iff nm.
