This answer doesn't address the stated questions (1), (2), and (3), but rather suggests a nice mathematical interpretation of the problem (generalized to more than two options). You have $n$ values $p_1, p_2, \ldots, p_n$ which add up to an integer sum $s$, in this case $s = 100$. You can divide each of them into integral and fractional parts $p_i = q_i + r_i$, where $q_i$ is an integer and $r_i$ is a real number in $[0, 1)$. Clearly $r_1 + r_2 + \cdots + r_n$ must be an integer, say $k$. The rounded form of each value is either $q_i$ or $q_i + 1$; that is, we will replace each $r_i$ with either zero or one. We are allowed only $k$ ones. Then the question is, which of the $p_i$ should get one of the $k$ ones, and which ones should not?
A natural way to do this is to define some function $f$ which tells how much each value "deserves" a one. Then you take the top $k$ values according to $f$ and give them ones (i.e. round them up) and give the rest zeroes (round them down). If you wanted to treat all values equally, you would set $f = q_i$. For example, if you had values $33.3\%$, $33.3\%$, and $33.4\%$, then $k = 0.3 + 0.3 + 0.4 = 1$, and the values of $f$ are $0.3$, $0.3$, and $0.4$, so you round as $33\%$, $33\%$, and $34\%$. (This is better than rounding each value independently, because then you get $33\%$, $33\%$, and $33\%$, and you've dropped a percentage point in the process.) If you want to give low values a boost, you could try $f = q_i + (s-p_i)/s$. Then a value of, say, $2.xx\%$ need only be at least $2.02\%$ to beat $50.5\%$ and $90.9\%$ in the Who Wants to Be Rounded Up competition.