First of all, it is **not** the case that if ZFC is consistent, then ZFC + CH + "There is a measurable" is consistent; a measurable cardinal has *much* greater consistency strength than ZFC alone. What **is** true is that if ZFC + "There is a measurable" is consistent, then so is ZFC + CH + "There is a measurable;" I think this is what you mean.

*A bit of clarification, just to pin things down: when we say "If $A$ is consistent, then $B$ is consistent," this is usually shorthand for "$T$ proves $Con(A)\implies Con(B)$" where $T$ is some reasonable weak theory (PRA is almost always enough); such an interpretation is needed to avoid stupidity (if ZFC is consistent, then the sentence "PA is consistent implies ZFC is consistent" is technically true, even though what we ***mean** by it clearly isn't). So, elaborating a bit on the previous paragraph, here are a couple things PRA proves:

*"If ZFC + "There is a measurable" is consistent, then ZFC + CH + "There is a measurable" is consistent."*

*"If PRA proves "If ZFC is consistent, then ZFC + "There is a measurable cardinal" is consistent," then ZFC is inconsistent."*

*No, that second one wasn't a typo.*

Now re: question 1, you ask:

how is it possible that ZFC + "The cardinality of the continuum is a real-valued measurable" implies $\neg$CH while ZFC+ "There exists a (two-valued) measurable cardinal" is consistent with CH since the two theories are equiconsistent?

There's no tension here - equiconsistent theories can disagree. Equiconsistency doesn't mean that they're consistent *together*, but rather that each is consistent if and only if the other is consistent. For example, ZFC+CH and ZFC+$\neg$CH are equiconsistent, but clearly inconsistent with each other.

So there's no issue.

Re: question 2, off the top of my head I'm not sure, but if memory serves Solovay showed that if there is a real-valued measurable, then there is an inner model with a measurable *(it is certainly true that he showed that "there is a real-valued measurable" and "there is a measurable" are equiconsistent over ZFC; but it's possible he build an inner model of a forcing extension, instead of an actual inner model, which would be a problem here)*. If my memory is correct, then the answer to 2 is "yes:" letting $M$ be that inner model, we have $M\models \vert\mathbb{R}^L\vert=\aleph_0$. But $L$ (hence $\mathbb{R}^L$) and $\aleph_0$ are absolute, so $V\models\vert\mathbb{R}^L\vert=\aleph_0$.

*Note that this is stronger than what you ask for in (2): the conclusion is that if there is a real-valued measurable ***at all**, regardless of its comparison with the continuum, then $\mathbb{R}^L$ is countable.

ZFC+ "$2^{\aleph_0}=\aleph_2$" implies $\neg CH$ whileZFCis consistent withCHsince the two theories are equiconsistent? $\endgroup$ – bof Sep 13 '17 at 4:25expectthem to have the same consequences for cardinal arithmetic. $\endgroup$ – Noah Schweber Sep 16 '17 at 18:02