how can you show a) Deterministic pushdown automata (DPDA) are more powerful than finite automata and less powerful than a non-determinstic pushdown automata?
(1) First, show that any language that can be accepted by a finite automaton can also be accepted by a deterministic pushdown automaton. Recall that any language accepted by a finite automaton is accepted by a deterministic finite automaton, and a deterministic pushdown automaton can simulate a deterministic finite automaton simply by doing nothing interesting with its stack. Next, show there is a non-regular language accepted by a DPDA. 0^n 1^n is a good candidate. Prove this language is non-regular using the pumping lemma or Myhill-Nerode theorem, thenshow the DPDA that pushes on 0s and then switches to popping on 1s works.
(2) First, note that NPDAs can accept any language accepted by DPDAs since all DPDAs are also NPDAs that happen not to make use of nondeterminism. Next, find a language that has an NPDA but no DPDA. Even-length palindromes over the alphabet {0, 1} might work. There is a simple NPDA for this that nondeterministically guesses when the first half of the input has been read and switches from pushing to popping. To show there is no DPDA is more challenging. Perhaps you could argue as follows: suppose there were a DPDA. Then, in any configuration of the DPDA, only one transition would be possible. If string w leads to an accepting state in the DPDA and empties the stack, x00 may lead either to an accepting or non-accepting state (since x00 either may or may not be an even-length palindrome). This is a contradiction, though, so our DPDA does not exist. The same argument fails for NPDAs, by the way, because there can be multiple paths through, so one failed choice means nothing.