What is an example of a Turing-recognizable infinite word, which is not Turing-decidable?
There's is a significant difference between "arbitrarily long" and "infinite".

As a simple example, an integer can have an arbitrarily great magnitude; formally, for every integer, there exists a larger integer (its successor), which in turn has a successor, and so on. But all integers are finite; $\omega \notin \mathbb{N}$. (Or, if you prefer, $\infty \notin \mathbb{N}$.)

Similarly, the Kleene star operator $A^*$ represents the concatenation of an arbitrarily large number of elements from $A$, but not an infinite number of elements of $A$. 

There is an interesting part of formal language theory which deals with sets of infinitely-long strings (&omega;-strings), but such strings cannot be produced by any regular expression. (You can use &omega;-regular expressions, which include the infinite repetition operator $A^\omega$.) But you might want to master the material you're currently studying before venturing onto Buchi automata, interesting though they may be.

To answer your questions:


No, there are undecidable languages consisting only of finite strings. Indeed, since such languages are undecidable, it is not necessarily even decidable whether they are finite sets. A classic example of an undecidable language is the set of (descriptions of) Turing machines which halt on every input. Note that every Turing machine (like every integer) has a finitely-long description, so the language is a set (an infinite set, in this case) of finite strings.
A Turing machine is under no obligation to read its entire input. Consider the language of &omega;-strings which start with an $a$. This is clearly a set of infinitely-long strings, but only a single character needs to be examined to determine whether a string belongs to the language.
Sure, why not. For example, the Turing machine trying to determine whether its input describes a Turing machine which halts on every input could simulate the described machine using every possible input, but that is going to go on forever even if the language describe a Turing machine which always halts.
See, for example, deterministic Buchi automata. Basically, an &omega;-language $L$ can be deterministic if there is a language $L' \subset Pref(L)$ of finite prefixes which can deterministically predict inclusion in $L$.