This is not possible, assuming $(f(A),w)$ means writing the code of $A$ and then writing input $w$. Context-free (and regular) languages have a pumping property, that is certain substrings can be repeated. I assume you have seen the concept. The length-bounds on the repeated parts are determined by the language you are interested in, and can be arbitrary large. The "universal PDA" on the other hand has its own fixed pumping constant that cannot match arbitrary large values in the coded languages. This argument however only works if we can enforce having to avoid pumping in the $f(A)$ part of the description. For regular languages that is not a big problem, for context-free we need a rather strong pumping result.
Sorry, no details here. Not enough space ....
There is a good reason for allowing the set of final (or accepting) states to be empty.
You probably know the standard proof that regular sets are closed under complement. Given a complete DFA $(Q, A, \delta, i, F)$ accepting the regular language $L$, the DFA $(Q, A, \delta, i, Q-F)$ accepts the complement of $L$. This is still perfectly true if $F = Q$, in which case $Q-F$ is empty.
In other words, if you were assuming that the set of final states is nonempty, the previous proof would not work in all cases, which would be a pity.
To answer your second question, the minimal DFA recognizing the empty language has one state. This state is initial but not final. The minimal DFA of its complement, the full language $A^*$ also has only one state, which is both initial and final.
A final remark. In the case of a NFA, not only the set of final states can be empty, but also the set of initial states can be empty.
Best Answer
Suppose the finite automaton makes transitions through states $q_0\to q_1\to q_2\to\ldots\to q_n$ in the process of accepting some string. Then you can arrange that your PDA does essentially the same thing, but instead of making state transitions, it makes the top (and only) stack symbol be $q_0\ldots, q_n$. Whenever the FA would have read input symbol $\sigma$ in state $q_a$ and made a transition to state $q_b$, the PDA instead will read input symbol $\sigma$ with $q_a$ on top of the stack and replace $q_a$ with $q_b$ on the stack. When the PDA reaches the end of the input, it can look to see if the top symbol on the stack represents an accepting state of the FA; if so, it pops it off, accepting by having an empty stack, and if not, it does nothing, rejecting the input.
Formally, say that the FA has input alphabet $\Sigma$, state set $Q$, accepting state set $A\subset Q$,and transition function $\delta:\Sigma\times Q\to Q$. Then define a PDA with input alphabet $\Sigma$, stack alphabet $Q$, state set $S=\{\bullet\}$ (there's your one state), and transition function $d:(\Sigma\cup\{\epsilon\})\times Q\times S\to Q^\ast\times S$ as follows: For $\sigma\in\Sigma$, take $d(\sigma,q,\bullet) = (\delta(\sigma, q), \bullet)$. For end-of-input, have $d(\epsilon, q, \bullet) = (\epsilon, \bullet)$ if $q\in A$ and $(q, \bullet)$ if not.