I don't like that asterisk notation for the operator norm, so I'll write instead of ; and I'll use for the sum of the absolute values of all the entries of A.
To show (i), use the fact that ( is the transpose of A), and check that . Then is a positive matrix, so its operator norm is equal to its largest eigenvalue. This is less than (or equal to) the sum of the eigenvalues, which in turn is the sum of its diagonal elements and therefore less than the sum of the absolute values of all its elements.
Putting all that together, you see that .
For (ii), use the Neumann series , which converges if .