I check it with the standard Garlerkin method and confirmed that it is right, in both cases (i) and (ii). 

Discribe the proof briefly (under (ii)): 

Take a CONS of $H$ as $h_1 = 1$ and $h_k(x) = \cos[(k-1)\pi x]$. 

First notice that if $V(u) = V^\dagger(\langle u, h_1 \rangle, \ldots, \langle u, h_N \rangle)$ and $\langle B(u), h_k \rangle = B_k^\dagger(\langle u, h_1 \rangle, \ldots, \langle u, h_N \rangle)$ for some $V^\dagger$, $B_k^\dagger \in C_b^2(\mathbb R^N)$ for $k \leq N$, and $B_k^\dagger = 0$ for $k > N$, then it becomes an $N$-dimensional diffusion and a infinite-dimensional OU process, and the result is obvious. 

Then take $V_N^\dagger$ and $B_{N, k}^\dagger$ be the marginal expectation, i.e., for $x \in \mathbb R^N$ and $k \leq N$, 
$$V_N^\dagger(x) = \int_{\mathbb R^\infty} V\left(\sum_{1}^N x_kh_k + \sum_{N+1}^\infty y_kh_k\right)\prod_{N+1}^\infty \Phi_k(y_k)dy_k, $$
$$B_{N, k}^\dagger(x) = e^{2V_N^\dagger(x)}  \int_{\mathbb R^\infty} U_k\left(\sum_{1}^N x_kh_k + \sum_{N+1}^\infty y_kh_k\right)\prod_{N+1}^\infty \Phi_k(y_k)dy_k, $$
where $U_k = e^{-2V}\langle B, h_k \rangle$ and $\Phi_k(x) = \sqrt{\lambda_k\pi^{-1}}e^{-\lambda_kx^2}$. This finite-dimensional approximation protect the condition (ii), so we have the invariant measure, due to the previous step. Finally take the limit.