No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant: $$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$ By the way, the invariant subspace problem for $UV$ is still open in full generality. https://mathscinet.ams.org/mathscinet-getitem?mr=353015 (Added: the Bishop operator wasn't exactly $UV$ and I'm not sure if my last comment was correct.)

No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant: $$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$ By the way, the invariant subspace problem for the Bishop operator $f(x)\mapsto xf(x+\theta)$ is still open in full generality. https://mathscinet.ams.org/mathscinet-getitem?mr=353015

No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant: $$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$ By the way, the invariant subspace problem for $UV$ is still open in full generality. https://mathscinet.ams.org/mathscinet-getitem?mr=353015

No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant: $$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$ By the way, the invariant subspace problem for $UV$ is still open in full generality. https://mathscinet.ams.org/mathscinet-getitem?mr=353015 (Added: the Bishop operator wasn't exactly $UV$ and I'm not sure if my last comment was correct.)