ok, here's a solution to 1):

using the fact that $a-1 < [a] \leq a,$ the equation 1) gives us: $63x - 6 < 12345 \leq 63x.$ therefore $[x] \in \{195,196 \}.$ now let $x=[x] + \theta,$ where $0 < \theta < 1.$ then 1) will give us:

$[2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=12345 - 63[x].$ so we must have $12345 \geq 63 [x],$ which doesn't hold for $[x]=196.$ hence we must have $[x]=195$ and therefore:

$[2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=60.$ using the inequality $[a] \leq a,$ we'll get: $60 \leq 62 \theta.$ thus: $\frac{30}{31} \leq \theta < 1.$ therefore $\frac{60}{31} \leq 2\theta < 2,$ and hence $[2 \theta]=1.$ also $\frac{120}{31} \leq 4\theta < 4$ and so

$[4 \theta]=3.$ similarly we see that $[8 \theta]=7$ and $[16 \theta]=15.$ thus: $26 + [32 \theta]= [2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=60,$ which gives us: $[32 \theta]=34,$ which is impossible because $0 < \theta < 1$

and hence $[32 \theta]$ is at most $31.$ so the equation has no solution!