If I have a value of $5.868709...×10^{−7}$, and an uncertainty of $7.88431...×10^{−12}$, is it correct to write this as $5.86871(8)×10^{−7}$ or $5.8687(8)×10^{−7}$?
A problem I have with the first is the values used to calculate the $5.868709...×10^{−7}$ were to five significant figures, so is it wrong for me to have six significant figures in my answer?
In the first place, I would write
$$ 5.868\,709\cdots×10^{−7} \pm 7.884\,31\cdots×10^{−12} $$
instead as
$$ (5.868\,709 \pm 0.000\,078\,8431 )×10^{−7} $$
with (a) grouped digits, (b) a common exponent, and (c) no ellipses. Next I would start removing “insignificant” digits.
The parenthesis notation gets rid of the leading zeros in the uncertainty. You should include the uncertain digits, rather than truncating or rounding them; think of them as “guard digits” for future computations, so that you can postpone as much rounding as possible for as long as possible. So the correct notation would be one of
\begin{align} &5.868\,709(79)&{}\times10^{-7} \\ &5.868\,71(8)&{}\times10^{-7} \end{align}
depending on whether you also keep a guard digit in your uncertainty. The Particle Data Group has a useful statement (section 5.3) about how many significant figures they include in their displayed uncertainties.
Going the other way, if you see a number like $\pi = 3.146(5)$, that means something like $3.141 < \pi < 3.151$, with appropriate caveats about confidence limits. Here you can see that rounding away the uncertain digit,
$$ 3.146 \pm 0.005 \not\to 3.15\pm0.005 $$
would erroneously exclude the correct value.
As far as having “gained” a significant figure: don’t fret about that. (Though you may be miscounting. The string $5\,868\,709$ has seven significant figures; moving the decimal point around using scientific notation doesn’t change that count.) The significant-figure approach is a heuristic for people who, for whatever reason, can’t or won’t keep track of uncertainties quantitatively. The usual rule is that, absent other information, the uncertainty in the least significant digit is one unit. So if you say,
“I trust $5.868\,709$ to five significant figures”
then I read instead
“the value is $5.868\,7(1)$, or maybe $5.868\,71(10)$, not to suggest those are terribly different”
You have “gained a digit” because your result, $5.868\,71(8)$ is has slightly smaller uncertainty than the naïve sig-fig approach.
Your uncertainty of $7.88431 \times 10^{-12}$ can be written as $0.0000788431 \times 10^{-7}$. But the value itself, $5.868709 \times 10^{-7}$ is known only to 6 decimal places, so the uncertainty cannot sensibly be given to more than 6 decimal places when expressed with the same $10^{-7}$ multiplier. So we could give the result as $$(5.868709±0.000079) \times 10^{-7}$$ I'd be more inclined to give your value as the safer $$(5.86871±0.00008) \times 10^{-7}.$$
