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Return to NFA to DFA.
A: I would say 1 because the $n$ states can be connected through $ε$ moves and hence all of these will be combines to a single start state in DFA.
In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example
<graphviz caption='NFA'> digraph example2 { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1; node [shape = circle];
""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }
<graphviz caption='DFA'> digraph example2 { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1; node [shape = circle];
""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }