a. while you can brute force this one in time, one simple trick to make it faster is to treat the symbols as bits and interpret the grid as numbers. It then becomes a matter of locating the reflection point.

b. It’s not much of a difference to solve b. The trick here is that if you did the bit stuff I suggested above, you’d quickly realise that a smudge interpreted as a binary number is a power of two. Finding a smudge is equivalent to if the bitwise XOR of two numbers is a power of 2, which can be done with some bitwise magic.