0.5×2=1 but 0.5+0.5=1.0. Explain? So when we multiply $0.5$ (approximate number) by $2$ (exact number), we get $1$, since our product must contain as many significant figures as $0.5$. 
When we add $0.5$ to $0.5$ (both approximate), we get $1.0$, since our sum must be as precise as the least precise of the above two numbers.
But aren't $0.5+0.5$ and $0.5×2$ the same thing? 
(I suspect it has something to do with both the $0.5$ being the same quantity. But it's only a hunch.)
 A: The difference is: in the addition case, you multiply with exactly 2. In the multiplication case, you multiply with approximately 2.
In the addition case, absolute errors add up, which means only the errors in 0.5 add up. You get potentially an error of magnitude $0.05+0.05=0.1$, in which case it makes sense to round to 1.0 (even though the actual result is anywhere between 0.9 and 1.1).
In the multiplication case, relative errors (approximately) add up, so if you have relative errors 10% (from 0.5) and 25% (from 2), your relative error is 35% and the result is (approximately, again) between 0.65 and 1.35, so you know very little about the next digit after 1, that is why it is approximated as just 1. (Try for youself, multiply $0.45 × 1.5$ and $0.55 × 2.5$ and see in which interval the result really fits.)
A: These are "rules of thumb" for numerical calculation, which are only approximately correct.   
For $0.5 \times 2$ we have:
$$
\text{if}\quad
\frac{45}{100} < a < \frac{55}{100}
\quad\text{and}\quad
b = 2
\quad\text{then}\quad
\frac{90}{100} < a \times b < \frac{110}{100}
\tag{1}$$
but the conclusion $1.0$ means
$$
\frac{95}{100}< a \times b < \frac{105}{100}
$$
but that conclusion is not (precisely speaking) justified.  
For $0.5 + 0.5$ we have
$$
\text{if}\quad
\frac{45}{100} < a < \frac{55}{100}
\quad\text{and}\quad
\frac{45}{100} < b < \frac{55}{100}
\quad\text{then}\quad
\frac{90}{100} < a + b < \frac{110}{100}
\tag{2}$$
but the conclusion $1$ means
$$
\frac{50}{100}< a + b < \frac{150}{100}
$$
so in this case the conclusion is much more than justified.  
Note that the conclusions in (1) and (2) are the same, as you asserted.  But the true state of that conclusion is somewhere between $1$ and $1.0$.
