Decide if a string is in a language without simulating the automata accepting the languge Is it possible for a Turing machine with input of a DFA that accepts a finite language and a string to decide whether the string is in the language without "fully simulating" the DFA on the string?

More formally, Is there a TM $M$ with a constant $C_M$ that can decide $L$, where for every input it halts at most $C_M$ steps after reading $w$'s last bit where $L$ is given by:

$L=\{\langle D,w\rangle \mid D\text{ is a DFA that accepts a finite language and $D$ accepts $w$}\}$?

I was thinking it is not possible, though I am not sure because maybe one can deduce from the encoding of the DFA without simulating the DFA what is the language.

I can't see how this can be done but not sure how to prove it can't be done.
You ask two questions.

For your first question, about whether a TM can do this without fully simulating the DFA:

The problem statement is not well-defined.  Given a TM, you can't tell whether it "fully simulates the DFA on a string" or not.  (That's undecidable.)  There are all sorts of ways that a TM might effectively "simulate", but in an obfuscated way that makes it hard for you to realize that's what it is effectively doing.

Your language $L$ is certainly decidable.  It has the form $L=L_1 \cap L_2$ where $L_1$ and $L_2$ are both decidable, hence $L$ is decidable too.

For the second question, about whether a TM can do this using at most a constant number of steps after reading the last bit:

Sure.  That's always possible, for any decidable language.  After reading all but the last bit, the TM can decide what it will do if the last bit is 0 and what it will do if the last bit is 1 (doing as much computation as necessary to handle both cases).  Then, when it reads the last bit, it will immediately be able to produce the correct output, using only a constant number of steps.  This doesn't really have anything to do with whether the TM is or isn't simulating the DFA, though.