Decidability of whether a language described by Turing machine is regular
This language/problem is neither decidable nor partially decidable (r.e). You can use Rice's theorem to prove it. The following proves that the language is not decidable.

"$L$ is regular" is a nontrivial property of recursively enumerable languages since there are regular and non-regular languages which are r.e. For example, $L_1 = \{0^{2n} \mid n &gt; 0\}$ is regular, while $L_2 = \{0^n1^n \mid n &gt; 0\}$ is not regular, but both are r.e. Hence, by Rice's theorem the language $\{\langle M \rangle \mid L(M) \text{ is regular } \}$ is not decidable.  

In order to prove that this language is not r.e. you could use Rice's theorem for recursively  enumerable index sets.