(s) regarding nucleophilic substitution and elimination reactions.

A is correct: $S_N2$ proceeds with inversion. In the substrate the priority order is $\mathrm{Br}>\mathrm{C_2H_5}>\mathrm{CH_3}>\mathrm{H}$ and in the product $\mathrm{I}>\mathrm{C_2H_5}>\mathrm{CH_3}>\mathrm{H}$, so the same ranking pattern holds and inversion changes $R$ to $S$. B is incorrect: the rate-determining step of an $S_N1$ reaction involves only the halide, so doubling $[\mathrm{OH^-}]$ leaves the rate unchanged. C is correct: in the anti-periplanar conformation of the meso isomer the two methyl groups are also anti, giving the trans alkene; the $dl$ pair gives $cis$-but-2-ene. D is incorrect: E2 requires an axial leaving group and an axial anti H. In the trans isomer the diaxial conformer places the methyl axial, so the hydrogen on that carbon is equatorial; the elimination goes towards the other side, giving 3-methylcyclohexene. Hence A and C.