Your statements are nearly correct, but none of them is fully
exactly right, in that there are in each case missing consistency
hypotheses. So let me explain how each of them can be improved in
some small way.

For the first part of statement (1), the correct thing to say is
that *if ZFC is consistent*, then ZFC does not prove Con(ZFC).
This is an immediate consequence of the 2nd incompleteness
theorem. For the second part, although it is widely believed that
ZFC is consistent, we don't actually *know* that ZFC might be
consistent; it is an extra assertion going beyond the ZFC axioms.
That is, the assertion that ZFC is consistent can legitimately be
made only under the assumption that it is true or under a stronger
assumption, such as the existence of large cardinals, which
implies that ZFC is consistent. Similarly, if ZFC really is consistent, then many would say it is wrong to say that it "might not be consistent", since that wouldn't be the case. However, a closely related correct statement is that if ZFC is consistent, then so is the theory ZFC+$\neg$Con(ZFC). That is, if ZFC is consistent, then it is consistent with ZFC to hold that it isn't consistent.

For the first part of the second statement, yes, one can prove in
ZFC that if there is an inaccessible cardinal, then Con(ZFC)
holds. This is simply because it is relatively easy to see that if
$\kappa$ is inaccessible, then the rank initial segment $V_\kappa$
satisfies all the ZFC axioms, and so there is a model of ZFC and
consequently Con(ZFC) holds.
The correct thing to say for the second part of statement (2) is
that *if ZFC is consistent*, then so is the theory
ZFC+$\neg$Con(ZFC+"there is an inaccessible cardinal"). This
follows from the incompleteness theorem applied to the theory
asserting that there is an inaccessible cardinal. Since ZFC+I, if
consistent, does not prove Con(ZFC+I), then either every model of
ZFC is a model of $\neg$Con(ZFC+I) or else there is a model of
ZFC+I+$\neg$Con(ZFC+I). Something a little closer to your
statement is that *if ZFC+Con(ZFC) is consistent*, then so is the
theory asserting ZFC+Con(ZFC)+$\neg$Con(ZFC+I).

Statement three is not true in the generality in which you make
it. For example, one of the things that you can prove in ZFC+I is
that 1=1, but it is not correct to say that this "never implies
that ZFC proves 1=1 in the actual natural numbers", since in fact
ZFC does prove that statement. But what you probably mean to say
instead is that the arithmetic consequences of ZFC+I is a strictly
larger theory than the arithmetic consequences of ZFC. This is
true, *provided that we assume ZFC is consistent*, since one of
the arithmetic consequences of ZFC+I is Con(ZFC) itself, which is
not provable in ZFC, provided that ZFC is consistent.