Need to find series of equation products from given reagents
I will assume that all of these reactions are happening in solution. Since we've established that all these reactions are redox reactions, we know the general structure of each reaction--- $$\text{reducing agent} + \text{oxidizing agent} \to \text{oxidized thing} + \text{reduced thing}$$ ---and we generally won't have to worry about other things (like what you've mentioned in your edit) happening. I will work out one reaction, and leave you with the rest. Let's look at reaction 4.

We can ignore $\ce{Na+}$ and $\ce{K+}$, because we know from prior experience that these are relatively inert spectator ions that won't participate in most reactions. Then we have the three species $\ce{S2O5^2-}$, $\ce{NO3^-}$, and $\ce{C12H22O11}$. 


$\ce{N}$ has its maximum oxidation state of $+5$ in $\ce{NO3^-}$, so we expect that $\ce{NO3^-}$ will be a good oxidizing agent. $\ce{NO3^-}$ is commonly reduced to $\ce{NO2^-}$ in solution.
On the other hand,  $\ce{S}$ has an oxidation state of $+4$ in $\ce{S2O5^2-}$ and a maximum oxidation state of $+6$, so we might expect $\ce{S2O5^2-}$ to be a reducing agent. Fully oxidized $\ce{S}$ is usually present as $\ce{SO4^2-}$ in solution.
Finally, $\ce{C}$ also has its maximum oxidation state of $\ce{+4}$ in $\ce{C12H22O11}$, so $\ce{C12H22O11}$ is likely to also be a reducing agent. Fully oxidized $\ce{C}$ is commonly present as $\ce{CO2}$.


This gets us the three half-reactions
$$\begin{aligned} \ce{NO3^- + H2O + 2e- &amp;-&gt; NO2^- + 2OH-}\\ \ce{S2O5^2- + 6OH- &amp;-&gt; 2SO4^2- + 3H2O + 4e^-}\\ \ce{C12H22O11 + 48OH- &amp;-&gt; 12CO2 + 35H2O + 48e-} \end{aligned}$$ where we've also assumed basic conditions and hence added $\ce{H2O}$ and $\ce{OH-}$ as required to balance out charges and atoms on both sides of each equation. Now we simply need to take linear combinations of these half-reactions to cancel out electrons on both sides, and we're done.



In response to your comment, everyone starts out bad at chemistry. You'll get better.