Solving SAT correctly on all but $poly(m)$ formulas
I'll provide a sketch.

Given any formula $\varphi$ of size $m$, we can rewrite it in an equisatisfiable way as $\varphi \land p$ where $p$ is a fresh propositional variable. In this way, we can generate an infinite amount of equisatisfiable formulas.

Now, exploiting this, it is easy to find a way to generate, in PTIME, $2 poly(m')+1$ formulas $\varphi'$ equisatisfiable wrt $\varphi$, where each such $\varphi'$ is of size $m'$. We can then decide SAT on $\varphi$ by taking what is the result of the majority of $\varphi'$ instances ($poly(m')+1$ instances), and return that. Indeed, among the $2poly(m')+1$ test, at most $poly(m')$ can have a wrong result, so the result of the majority must be the correct one.

This runs a PTIME algorithm a polynomial number of times ($2 poly(m')+1$, where $m'$ is polynomial wrt $m$), so we indeed solve SAT in PTIME.

To generate all the needed $\varphi'$, we note that the "longest" variable name in $\varphi$ can't more than $m$ bit long for obvious size reasons. Hence if we take $p$ to be any variable whose name has size $m+1$ bits it is surely fresh, and this makes $\varphi'$ to have size $m' = m+(m+1)+O(1)$. Further, we have $2^{m+1}$ choices of such $p$, which is (asymptotically) a large enough space to choose $2 poly(m')+1$.