The remainder when $1$ is divided by $2$ is $1$, since $1=(0)(2)+1$ and $0\le 1\lt 2$.
In general, if $0\le a\lt m$ then $a\operatorname{\%}m=a$.
In general, when you divide an integer $a$ by a positive integer $m$, there is a quotient $q$ and a remainder $r$. So
$$a=qm+r,$$
where $0\le r\lt m$.
For instance, if $a=30$ and $m=12$, then $q=2$ and $r=6$. If $a=5$ and $m=12$, then $q=0$ and $r=5$.
In the case where $a=1$ and $m=2$, the quotient is $0$ and the remainder is $1$.
Remark: It is useful to have concrete images to go along with more abstract descriptions. Suppose that we have a box that contains $a$ cookies, and we have $m$ kids in the room. We give a cookie to everyone (if we can). Then we do it again, and again, doing a full round each time. The number of cookies left in the box is the remainder when $a$ is divided by $m$, it is what's left over.
For example, if $a=40$ and $m=12$, we do $3$ full rounds, each kid gets $3$ cookies. This $3$ is called the quotient. We will have $4$ cookies left over, the remainder is $4$, in symbols $40\operatorname{\%} 12=4$. If we start with $72$ cookies, the remainder is $0$.
But if we start with $5$ cookies, then we can't even get started, we cannot distribute cookies without causing a riot. So the quotient is $0$, nobody gets a cookie. And all the cookies are left over, the remainder is $5$, that is, $5\operatorname{\%}12=5$.
If you compute $1/3$ and store it in $Q$, and then compute $3*Q$, you'll find that you don't get 1 -- there's a tiny numerical imprecision, related to the fact that calculators use binary, in which $1/3$ is not representable in a finite number of bits(just as it is not in decimal, where it cannot be represented with a finite number of decimal digits: you have to write $.33333\ldots$).
The same is true for $\log -1$: it can't be represented exactly in binary. So when you exponentiate it, you get something very near $-1$; the error in this case happens to be in the imaginary component rather than the real part, but it's the same phenomenon.
By the way, 1+2e-13i means $-1 + 2 \times 10^{-13} \Bbb {i}$.
Best Answer
The calculator automatically uses scientific notation for numbers smaller than $0.001$ in absolute value, no matter what your mode settings are; this cannot be overridden.
The most straightforward way to get the calculator to immediately tell you how many zeroes there are after the decimal, in such a case, is to add $1$:
Just ignore the $1$ in front of the decimal point and you have the answer you want.
Note that in general, this will lose you precision:
In this case, it's not too bad: we can use the second output to count zeroes and the first for more digits. Sometimes, it will lose you all of the precision:
But in this last example, maybe you don't want the number as a decimal after all.