How do these cascade reactions work?
how do you do the others?


You do it like a detective. Even the slightest details matter, and you've to link a lot of concepts together.

The main theme you're missing in your question is this equilibrium:

$$\ce{ CrO4^2- &lt;=&gt;[\ce{H+}][\ce{OH-}] Cr2O7^2- }$$

Dichromates are supposed to be stable in an acidic medium. So using this you can figure out that ($7$) is $\ce{H+}$ and ($8$) is $\ce{OH-}$. Again for ($5$) you're adding a base which is converting it to a chromate. So ($5$) should be a dichromate $\ce{Cr2O7^2-}$.

So we're only left with ($9$) and $\ce{Cr^3+}$ and they are bound to go with each other. Since this is not an exam here's a reasoning too,

$\ce{K2Cr2O7}$ is a strong oxidizer in acidic mediums. If it's going to oxidize others, it will itself get reduced. To be precise here's the reaction that's taking place:

$$\ce{Cr2O7^2- + 14H+ + 6e- -&gt; 2Cr^3+ + 7H2O}$$