Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have non-integer solutions? Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have any non-integer solutions? If so, how do you find them?

I'm assuming your $x$ and $y$ are real numbers. Now, we can write $x= \lfloor x \rfloor + \{x \} $ and $ y=\lfloor y \rfloor + \{y\} $ where the $\{ \}$ here denotes the fractional part. We have that $\{x\} \neq 0$ and $\{y\} \neq 0$ since $x$ and $y$ are not integers. Your equation then reads
$$ \lfloor x \lceil y \rceil \rfloor = \lceil x \lfloor y \rfloor \rceil $$
$$ \lfloor (\lfloor x \rfloor  +\{x\})(\lfloor y \rfloor +1) \rfloor = \lceil (\lfloor x \rfloor  +\{x\}) \lfloor y \rfloor \rceil   $$ which gives
$$ \lfloor x \rfloor( \lfloor y \rfloor +1) + \lfloor \{x\}(\lfloor y \rfloor+1) \rfloor = \lfloor x \rfloor \lfloor  y \rfloor + \lceil \{x\}\lfloor y \rfloor \rceil $$ which after simplification and rearranging becomes
$$ \lfloor x \rfloor = \lceil \{x\}\lfloor y \rfloor \rceil - \lfloor \{x\}(\lfloor y \rfloor+1) \rfloor  $$ 
Now, $$ \{x\}\lfloor y \rfloor < \{x\}(\lfloor y \rfloor+1) < \{x\}\lfloor y \rfloor +1 $$ which means that after rounding the small one up and the big one down they either become equal or the one rounded up is one more than the other. Hence, we see that $ \lfloor x \rfloor = 0,1$. If $ \lfloor x \rfloor = 0$ we have that $x$ is between 0 and 1 and that $\{x\}=x$ and the rounded up and down are equal, so there is an integer between $x\lfloor y \rfloor$ and $x\lfloor y \rfloor + x$. Call it $n$. Then $$ x\lfloor y \rfloor \leq n \leq x\lfloor y \rfloor + x $$ giving $$ \frac{n}{x} -1 \leq \lfloor y \rfloor \leq \frac{n}{x} $$ What does this mean? It means, choose an $x$ between 0 and 1, and an integer $n$. Compute $ \frac{n}{x}$ and $\frac{n}{x} -1 $ and find an integer between them. In general, there will only be one integer between them, unless $\frac{n}{x}$ is an integer ( so $x$ is rational ), in that case both the upper and lower bounds are integers and there are two possibilities. This integer you find is $\lfloor y \rfloor$ - but what is $y$? It doesn't matter, as long as $ \lfloor y \rfloor$ is the right value - so add any fractional part to $y$ that you wish. As for the case $\lfloor x \rfloor = 1$ ... how about I leave that as an exercise for you.
