ok, here's a solution to 1):

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

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

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

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

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