Oracle comparing $EXP$ with $UP$

Heller (Theorem 6) gave an oracle relative to which $NP=EXP$, and Homer & Selman gave an oracle relative to which $P=UP$ and $\Sigma_2^P=EXP$.

Beigel, Buhrman, Fortnow (freely available author's version) gave an oracle in which $P=\oplus P$ and $NP=EXP$ holds.

1. Is there an oracle that gives $P\neq UP=coUP=NP=coNP=\oplus P=EXP$?

2. Is there an oracle that gives $P=UP=coUP\neq NP=coNP=\oplus P=EXP$?

Or is it known $UP\neq EXP$ unconditionally?

$\mathsf{UP} \neq \mathsf{EXP}$ is open. A UP-generic oracle* should make $\mathsf{P} \neq \mathsf{UP} = \mathsf{EXP}$, and since $\mathsf{UP} \subseteq \mathsf{\oplus P} \subseteq \mathsf{EXP}$ relative to any oracle, this should resolve 1. (I say "should" because I haven't checked all the details...)

*UP-generic oracles are discussed, for example, by Fenner-Fortnow-Kurtz-Li (author's freely available version) and also (shameless self plug) a quick tutorial in Section 5.1 here (author's free version)

• Since your asnwer was earlier I will accept your answer. – T.... Aug 8 '17 at 2:51
• But what is an UP generic oracle? – T.... Aug 8 '17 at 7:35

The quoted Beigel, Buhrman, and Fortnow paper gives a solution to 2 in Theorem 1.8: there is an oracle relative to which $\mathrm{P=Mod_3P}$ (which implies $\mathrm{P=UP}$), and $\mathrm{\oplus P=NP=EXP}$ (which, together with the first equality, actually implies $\mathrm{EXP=ZPP}$).