How to calculate the amount of water produced by the incomplete combustion of propane?
Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:
$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -&gt; $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\
\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}
\end{multline}$$
This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.
The above case reduces to your formula for $x=5$ and $y=0$.



For the second question, your equation has a small error, correct is:
$$\ce{C3H8 (g) + 5O2 (g) -&gt; 3CO2 (g) + 4H2O (g)}$$
But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:
$$\ce{C3H8 (g) + 2O2 (g) -&gt; C (s) + 4H2O (g)}$$
So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.

You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the ideal gas law:
$$pV =n\mathcal{R}T$$
You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.

You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.


   $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$


From the reaction equation you can now determine the amount of substance of water, that will be produced.


   $n(\ce{H2O})\approx 2.8~\mathrm{mol}$


You can convert that easily into a mass now.


   $m(\ce{H2O})\approx 50~\mathrm{mol}$


There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.
You are of course right, that you will also have a rise in temperature.

Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.