At least one of (a) or (b) is true unless $x + y$ is an integer, but $x$ is not an integer. Exactly one of (a) and (b) is true if none of $x$, $y$, $x + y$ are integers. Both (a) and (b) are true if at least one of $x$, $y$ is an integer.

Method: for $x$ a real number, write $x = [x] + \{x\}$, where $0 \le \{x\} < 1$ is the fractional part of $x$. Then (a) is equivalent to $\{x+y\} = \{x\} + \{y\}$, which in turn is equivalent to $\{x\} + \{y\} < 1$. Likewise, equivalent to (b) is the statement: $x$ or $y$ is an integer or $\{x\} + \{y\} > 1$. This relies on the fact that $\{-x\} = 1 - \{x\}$ if $x$ is not an integer.

The best way of visualising the solution is as follows. Notice first that adding an integer to $x$ or to $y$ does not change whether or not $(x, y)$ satisfies (a) or (b). So we can replace $x$ by $\{x\}$ and $y$ by $\{y\}$. So now we are looking at all points $(x, y)$ in the unit square $\{(x, y) \mid 0 \le x, y < 1\}$. Check first that (a) is satisfied by all points in the lower triangle bounded by the diagonal $x+y=1$, plus the endpoints of the diagonal $(1, 0)$ and $(0, 1)$. Now $x \to 1-x$ maps the square onto itself. The $(x, y)$ satisfying (b) are those points $(1-x, 1-y)$ satisfying (a). By the above, these are the $(x, y)$ in the upper triangle bounded by $x+y=1$, plus $(1, 0)$ and $(0, 1)$. The only points in the square satisfying neither (a) nor (b) are those on the diagonal $x+y=1$, apart from $(1, 0)$ and $(0, 1)$ i.e. when neither $x$ nor $y$ is an integer but $x+y$ is an integer.