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In the peracid there are two oxygen atoms with oxidation state 0, so those are electrophilic. The ensuing reaction is called Prilezhaev epoxidation. In a single step, one oxygen atom is transferred from the peracid to the alkene, yielding the carboxylic acid and an epoxide (so it's the correct answer to the question). For the transition state, look it up on Wikipedia. (Tip for learning named reactions: learn them with the name that describes what the reaction actually does (provided there's one), i.e. Prilezhaev epoxidation instead of just Prilezhaev reaction, Wittig olefination instead of Wittig reaction, etc.)
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OsO4 is electrophilic because Os is in oxidation state 8, the reaction that follows is a [3+2]-cycloaddition of OsO4 at the alkene. This intermediate is hydrolyzed to yield an 1,2-Diol (i.e. not the epoxide pictured). Look up dihydroxylation for details. Keep in mind that in practice OsO4 is used in catalytical amounts and a cooxidans is used (because OsO4 is expensive and toxic).
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Ozone is the electrophile because of oxygen with oxidation state zero. Again, [3+2]-cycloaddition happens. The resulting compound (5-ring with 3 oxygen atoms in a row) rearranges (mechanism depends on solvent) to yield another 5-ring (only 2 oxygen in a row). The reductive workup yields two ketones/aldehydes (in this case: an aldehyde and formaldehyde, again: not the epoxide pictured). Look up ozonolysis for details.