Decidability of Turing machines and misconceptions on the halting problem
The problem of deciding whether, given a Turing machine $T$ and a word $w$, $T(w)$ halts is undecidable.
If a Turing machine $T$ is fixed, the problem of deciding whether, given a word $w$, $T(w)$ halts might or might not be decidable (depending on the choice of $T$).
As an example in which this problem is decidable consider the trivial Turing machine $T$ that immediately halts.
As an example in which this problem is undecidable consider an universal Turing machine $T$ in which $w$ encodes a pair $\langle M, w'\rangle$ where $M$ is some other Turing machine, and $w'$ is $M$'s input. The machine $T$ simply simulates $M'$ on $w'$. Clearly if we could solve the halting problem for this fixed $T$ and any $w$, we could also solve it in the general case.
If a word $w$ is fixed but the machine $T$ is part of the input of the problem, then the problem is undecidable regardless of $w$.
Suppose towards a contradiction that this problem was decidable, then given a Turing machine $M$ and a word $w'$ as input, you could decide whether $M(w')$ halts. To do so you construct a new Turing machine $M'$ that clears its tape, writes $w'$, and then simulates $M$ (an encoding of such a machine $M'$ is computable given $M$ and $w$). Clearly $M'(x)$ halts iff $M(w')$ halts, regardless of $x$. In particular, for $x=w$,  $M'(w)$ halts iff $M(w')$ halts. It then suffices to decide whether $M'(w)$ halts, which we can do by our hypothesis.
If both a Turing machine $T$ and a word $w$ are fixed, then the problem does not make much sense since you do not have a class of instances anymore... Notice that, in this case, either $T(w)$ halts or it does not, meaning that one of the following two Turing machines must answer correctly: 1) Always halt and accept. 2) Always halt and reject.