On $NP$, $\oplus P$ and $PP$?
Concerning 3, I believe $\mathrm{PP\subseteq\oplus P^{C_=P}}$, as there are at least $a$ numbers $x&lt;2^n$ satisfying $P(x)$ if and only if the number of $y&lt;2^n$ such that $P(y)\land|\{x\le y:P(x)\}|=a$ is odd. (Note that there is always at most one such $y$. That is, the argument actually shows $\mathrm{PP\subseteq UP^{C_=P}}$. (In fact, it even shows $\mathrm{UP^{PP}=UP^{C_=P}=UP^{C_=P[1]}}$.))