By the conditions, we know that $f(x+n) = 1 - f(x)$ when $n$ is odd and $f(x+n) = f(x)$ when $n$ is even.
Note that $a = \log_2 3 \in [1, 2)$, $2a = \log_2 9 \in [3, 4)$, $3a = \log_2 27 \in [4, 5)$. Therefore,
$$\text{f(a) + f(2a) + f(3a) = 1 - f(a - 1) + 1 - f(2a - 3) + f(3a - 4) = 2 - f ( 2 3 2 ) - f ( 2 9 8 ) + f ( 2 27 16 ) = 2 - ( 3 2 - 2 3 2 ) - ( 9 8 - 2 9 8 ) + ( 27 16 - 2 27 16 ) = 17 16 + ( 2 3 2 + 2 9 8 - 2 27 16 ) = 17 16 .}$$