2 Questions regarding Relative Consistency Proofs First Question:
Let IC be the statement "There is an inaccessible cardinal."
I have read that one cannot prove (in ZFC) the relative consistency of ZFC + IC w.r.t. ZFC.
i.e. $ Con(ZFC) \rightarrow (ZFC \nvdash Con(ZFC) \rightarrow Con(ZFC + IC)) $
The argument is as follows:
Assume $ ZFC \vdash Con(ZFC) \rightarrow Con(ZFC + IC) $
since $ ZFC + IC \vdash Con(ZFC)$
it follows that $ ZFC + IC \vdash Con(ZFC + IC)$ which contradicts Gödel's 2nd Incompleteness Theorem.
But from Gödel's Theorem we only can conclude $ \neg Con(ZFC + IC) $, so why is this a contradiction?
Second Question:
I call an $\in $-formula $\phi$ independent, if we have:
$ \Phi_{1,1} := Con(ZFC) \rightarrow Con(ZFC + \phi)$ and $ \Phi_{1,2} := Con(ZFC) \rightarrow Con(ZFC + \neg \phi)$
Now $ \Phi_{1,1}$ and $\Phi_{1,2}$ are again $\in $-formulas and could again be independent if we have:
$ \Phi_{2,1} := Con(ZFC) \rightarrow Con(ZFC + \Phi_{1,1})$ and $ \Phi_{2,2} := Con(ZFC) \rightarrow Con(ZFC + \neg \Phi_{1,1})$
In this case $\phi$ would not be decided by ZFC, but we could not know about this result. Can such a situation occur?

The Second Incompleteness Theorem says that if $T$ is a recursively axiomatized, consistent, and "strong enough" theory, then $T\not\vdash \mathcal{Con}(T)$.
Since ZFC+IC is recursively axiomatized and strong enough, then the Second Incompleteness Theorem gives us
$$ \tag{*} \mathcal{Con}(\mathsf{ZFC+IC}) \to \Bigl[ \mathsf{ZFC+IC}\not\vdash \mathcal{Con}(\mathsf{ZFC+IC}) \Bigr]$$
The argument you're quoting implicitly assumes that ZFC is not only consistent but true (at least as regards arithmetic), such that if ZFC proves $\mathcal{Con}(\mathsf{ZFC})\to\mathcal{Con}(\mathsf{ZFC+IC})$ then it must be that ZFC+IC is actually consistent. Thus by (*) we have $\mathsf{ZFC+IC}\not\vdash \mathcal{Con}(\mathsf{ZFC+IC})$ which contradics the earlier conclusion that $\mathsf{ZFC+IC} \vdash \mathcal{Con}(\mathsf{ZFC+IC})$.
