For A, given $\langle M\rangle$, simply run $M$ on $010$.
If $M$ accepts, accept.
If $M$ rejects or loops, the recognizer may reject or loop. Therefore the language is recognizable.
$\boxed{\mathrm{A}\in\mathrm{RE}}$
For B, nondeterministic TMs and deterministic TMs have the same recognition power. A deterministic TM can systematically simulate the computation tree of the NTM.
Hence,
$\boxed{\mathrm{B}\in\mathrm{RE}}$.
For C, notice the phrase:
$M\ \mathrm{does\ not\ accept}\ 101$.
This includes two possibilities $: \mathrm{reject}$ or $\mathrm{loop\ forever}$.
If $M$ loops on $101$, we cannot wait for some finite event that confirms that it will never accept. Thus this language is not Turing-recognizable.
For D, checking whether a TM accepts every possible string cannot be recognized simply by running strings one after another, since some simulations may loop forever.
Therefore,
Answer : $\boxed{\mathrm{A\ and\ B}}$
