The third language defines a constant Upper bound for the calculation steps of the machine.
There is no constant upper bound on the number of steps of $M$ in $L_3$. $L_3$ is the language of all machines that halt on at least one word $w$ while performing less steps than the length of $w$. The word $w$ is not fixed, all that is needed to $M$ to be in $L_3$ is for such a word to exists. The length $w$ must be finite by might be arbitrarily large.
$L_3$ is not in $R$. If $L_3$ was in $R$ then you wold be able to decide the halting problem. Given a given Turing machine $T$ and an input $w$, construct a Turing machine $T'$ that simulates $T$ on input $w$ and then accepts. If $T$ halts then it must do so within a certain number $x$ of steps, showing that $T' \in L_3$ since we can pick $w = 1^{x+1}$. If $T$ does not halt, then $T'$ also does not halt showing that $T' \not \in L_3$.
While the second language defines a Upper bound according to the word and may enter an infinite loop. (is it true ?)
I don't know what you mean by "may enter an infinite loop". It makes no sense to say that a language "enters an infinite loop". Moreover, a Turing machine $M$ in $L_2$ might or might not enter into an infinite loop, depending on its input. However you know for sure that there is at least one input to $M$ such that $M$ does not enter an infinite loop (since it must halt in less than $|M|$ steps).
Notice that, as before, there is no constant upper bound on the number of steps. Rather the maximum number of steps is determined by the length of the encoding of $M$ (which must be finite but can be arbitrarily large).
$L_2$ is in $R$ since there are at most $|M|-1$ different positions of the input word that can be read by a Turing machine $M$ that halts in less than $|M|$ steps. You can then simulate (all the possible execution paths of) $M$ for $|M|-1$ steps on all possible input words of length at most $|M|-1$ (there are only $|\Sigma|^{|M|-1}$ such words).
In addition, what is the difference between the first language and the second language?
If you meant the intuitive difference in the definitions of $L_1$ and $L_2$, then in the first language the number of steps by which $M$ must halt is specified as part of the word in $L_1$. Alternatively, you can think of it as being written in unary as a part of the input of a TM that decides $L_1$.
In contrast, the second language specifies the number of steps by which $M$ must halt using the length of the encoding of $M$ itself.
If you meant the set difference then $L_1 \setminus L_2 = L_1$.
$L_1$ is in $R$. since it suffices to simulate all the (possibly exponentially many w.r.t. $t$) execution paths of $M$ of length at most $t$.