Which languages, decided by a turing machine are decidable?
Nice question. 

Notations and terms

$M$ or $N$ means a Turing machine (TM), whose specification may or may not given. $\langle M\rangle$ is the description of $M$ according to a predefined effective encoding scheme for TMs.

$L(M)$ is the language recognized by $M$, i.e., the set of words accepted by $M$. At least that is what I have seen everywhere.

Whether a language is decidable or a language is decided by a TM is an entirely different although closely related concept. Let me quote the definition in the book introduction to the theory of computation by Michael Sipser. You could take a look at its definition at Wikipedia as well.


  We prefer Turing machines that halt on all inputs; such machines never loop. These machines are called deciders because they always make a decision to accept or reject. A decider that recognizes some language also is said to decide that language.
  
  DEFINITION 3.6. Call a language Turing-decidable or simply decidable if some Turing machine decides it.


Note that if $M$ is a decider, then $M$ decides $L(M)$. If it is said that a Turing machine $N$ decides some language $A$, then $N$ must be a decider and $A=L(N)$.

Application of Rice's theorem

Here is the Rice's theorem. 


  Let $P$ be any nontrivial property of the language of a Turing machine. Then whether a given Turing machine’s language has property P is undecidable.
  
  In more formal terms, let $P$ be a language consisting of Turing machine descriptions where $P$ fulfills two conditions. First, $P$ is nontrivial—-it contains some, but not all, TM descriptions. Second, $P$ is a property of the TM's language—-whenever $L(M_1) = L(M_2)$, we have $\langle M_1\rangle \in P$ iff $\langle M_2\rangle \in P$. Here, $M_1$ and $M_2$ are any TMs. Then $P$ is an undecidable language.


a) $P=\{\langle M \rangle\mid L(M)\subseteq 0^*\}$. 


WLOG, let 1 be an input alphabet symbol other than 0. $P$ is nontrivial since $P$ contains the description of a Turing machine whose language is $\{0\}$ but $P$ does not contain the description of a Turing machine whose language is $\{1\}$.
$P$ is a property of the TM's language—-whenever $L(M_1) = L(M_2)$, we have $L(M_1)\subseteq 0^*$ iff $L(M_2)\subseteq 0^*$

By Rice's theorem, $P$ is undecidable. 


b) $P=\{\langle M \rangle\mid L(M)\text{ contains at least one word of even length}\}$. 


$P$ is nontrivial since $P$ contains the description of a Turing machine whose language is $\{00\}$ but $P$ does not contain the description of a Turing machine whose language is $\{0\}$.
$P$ is a property of the TM's language—-whenever $L(M_1) = L(M_2)$, we have $L(M_1)$ contains at least one word of even length iff $L(M_2)$ contains at least one word of even length.

By Rice's theorem, $P$ is undecidable. 


c) By the definition of semi-decidable, a.k.a Turing-recognizable, $L(M)$ is semi-decidable for any $M$. Hence $\{ \langle M \rangle \mid L(M)$$\text{ is semi-decidable}\}$ contains all TM descriptions, which is a decidable language.

d) $P=\{\langle M \rangle\mid L(M)\text{ is decidable}\}$. 


Let $R=L(D)$ be a decidable language. For example, we can take $R=\{0\}$ and $D$ is a Turing machine that accepts 0 and rejects all other inputs. $\langle D\rangle\in P$. Let $H=L(N)$ be a semi-decidable but not decidable language. For example, we can take $H$ to be the language of the halting problem. Then $\langle N\rangle\not\in P$. Hence $P$ is nontrivial. 
$P$ is a property of the TM's language—-whenever $L(M_1) = L(M_2)$, we have $L(M_1)$ is decidable iff $L(M_2)$ is decidable.

By Rice's theorem, $P$ is undecidable. 


 For case a), if we restrict the input alphabets of all Turing machines to be the unary set $\{0\}$, then, of course, for every $M$, $L(M)\subseteq 0^*$. Then $\{ \langle M \rangle \mid\text{0 is the only tape alphabet symbol and } L(M) \subseteq 0^*\}$ contains all TM descriptions. It is decidable.