finding the parity(even or odd) of modulo calculation let a = b mod m, where b is 2 n bits long, and m is n bits long， n maybe large.
How to get the parity of a without know a. Because we don't want to know a, there maybe a quick algorithm to get the parity of a, in the perspective of theory of information.
Similarly, let a = b * c mod m, where b and c and m are n bits long.
How to get the parity of a without evaluating multiply and modular.
 A: You are making an unwarranted assumption that calculating the parity of $a$ must be easier than calculating $a$.  Here is a lower-bound argument which  refutes the premise underlying your question.
Let $m\ge 3$ be any odd number (user296602 has already explained why $m$ even is trivial), and choose $b$ to any long number consisting of digits between $1$ and $8$ such that $b$ is divisible by $m$.
[The existence of arbitrarily large $b$ of this type is a nice exercise but not really critical to this argument since for any given $m$ there are plenty of examples available (in the simplest case where $5 \nmid m$, we may simply take $b$ to be a string of repeating $3$s of an appropriate length, and in the general case we can appeal to an argument similar to For every positive number $n$, there exists a $n$ digit number having all odd digits and divisible by $5^n$
Let us add one additional constraint on $m$, that it not be a power of $5$.  If $m = 5^r$ then $b$ mod $m$ can be computed from just the last $r$ digits of $b$, which makes the question moot as it is already easy to compute $a$ exactly.
So far, we know that $a = 0$ so that $a$ is even.  But I claim there is no way for us to compute this without having examined every single digit of $b$.  Suppose we neglected to examine any single digit of $b$, say at position $d$.  Then we could replace $b$ by $b'$ where $b$ and $b'$ are identical except at position $d$, and the output of our algorithm would have to be the same.
By our choice of $b$, we may freely add or subtract one from position $d$.  Thus we may take $b' = b \pm 10^k$, where we don't control the value of $k$ (the argument needs to work for arbitrary $k$) but we do control the $\pm$ sign.
Since $m$ is divisible by at least one prime other than $2$ or $5$, $10^k$ is not divisible by $m$, so it has a non-zero residue $c$ mod $m$.  
If $c$ is odd, then $b' = b + 10^k$ has an odd residue mod $m$, which means we must examine that particular digit of $b$ in order to know the parity of $a$.
If $c$ is even, then we take the opposite sign: $b' = b - 10^k$ has residue $m-c$ mod $m$, which is odd.  And again this proves we must examine that digit of $b$ in order to know the parity of $a$.
We conclude that it is impossible to know the parity of $a$ without at least checking every digit of $b$, and in roughly that same amount of time we could have computed $a$ itself.
