The procedure I have to take is to first identify that both U(X) are unbiased for theta, and that U2(X) = E[U1(X)|T2(X)].
My question is How would that relate to U2 having a smaller variance than U1, and how are U2 and U1 related to being the minimal sufficient and sufficient statistic?
1)prove unbiased $U_1$ and $U_2$
$E(U_1)=E(E(U|T_1))=E(U)$
$E(U_2)=E(E(U|T_2))=E(U)$
2) prove $U_2=E(U_1|T_2)$
$T_2$ is minimal sufficient so it is sufficient and a function of any sufficient estimator like $T_1$ (by defination of minimal sufficient) so $T_2=f(T_1)$ and hence $\sigma(T_2) \subset \sigma(T_1)$
$E(U_1|T_2)=E\bigg(E(U|T_1)|T_2\bigg)=E\bigg(E(U|\sigma(T_1) )|\sigma(T_2) \bigg)$
By Tower property since $\sigma(T_2) \subset \sigma(T_1)$
$E\bigg(E(U|\sigma(T_1) )|\sigma(T_2) \bigg)=E(U|\sigma(T_2))=E(U|T_2)=U_2 $ so
$E(U_1|T_2)=U_2 $
3) prove $V(U_2) \leq V(U_1)$
$E(U_1|T_2)=U_2 $ so
$V(U_1)=E\bigg(V(U_1|T_2)\bigg)+ V\bigg( \underbrace{E(U_1|T_2)} \bigg)=E\bigg(V(U_1|T_2)\bigg)+V(U_2)$ so
$V(U_1)=E\bigg(V(U_1|T_2)\bigg)+V(U_2)$ hence
$V(U_1) \geq V(U_2)$