$f_i(w) = $ maximum value obtainable with $i$ objects and a knapsack capacity $W$.
$f_i(w) = \text{max}(f){i-1} (w) , \: v_i +f_{i-1}(w-w_i))$
$f_i$ depends on $f_{i-1}$ subproblem only.
So, statement I is correct.
For $f_i(x)$
$i \in \{0, 1, 2, \dots , n\}$
$x \in \{0, 1, 2, \dots, n \}$
Number of distinct subproblems $=O(nW)$
So, statement II is correct.