It seems that Betelgeuse is a lot closer, but even with its gargantuan size, it's still a lot smaller than Eta Carinae.
I'm pretty sure Betelgeuse will be a lot brighter. It is more than 10 times closer than Eta Carina. Because of the inverse square law, if they were the same intrinsic brightness, Betelgeuse would be more than 100 times brighter. As you say, Eta Carina is perhaps as much as 10 times more massive, but even if it is intrinsically 10 times brighter, it would still be 1/10 as bright as Betelgeuse.It seems that Betelgeuse is a lot closer, but even with its gargantuan size, it's still a lot smaller than Eta Carinae.
As a single star, a star originally around 150 times as massive as the Sun would typically reach core collapse as a Wolf–Rayet star within 3 million years. At low metallicity, many massive stars will collapse directly to a black hole with no visible explosion or a sub-luminous supernova, and a small fraction will produce a pair-instability supernova, but at solar metallicity and above there is expected to be sufficient mass loss before collapse to allow a visible supernova of type Ib or Ic. If there is still a large amount of expelled material close to the star, the shock formed by the supernova explosion impacting the circumstellar material can efficiently convert kinetic energy to radiation, resulting in a superluminous supernova (SLSN) or hypernova, several times more luminous than a typical core collapse supernova and much longer-lasting. Highly massive progenitors may also eject sufficient nickel to cause a SLSN simply from the radioactive decay. The resulting remnant would be a black hole since it is highly unlikely such a massive star could ever lose sufficient mass for its core not to exceed the limit for a neutron star.