The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form
$$(\text{number})(\text{unit}) = (\text{other number})(\text{other unit})$$
But of course, you can divide both sides of any equation by the same thing, and the equation will continue to hold. So you can also write the conversion factor as
$$\frac{(\text{number})(\text{unit})}{(\text{other number})(\text{other unit})} = 1$$
or
$$\frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})} = 1$$
Same equation, just rearranged a bit.
Now let's see what happens when you have some value, expressed in $(\text{unit})$, that you want to convert to $(\text{other unit})$. As you know, you can always multiply anything by 1, and that doesn't change the value at all. So, you can go look up the conversion relationship between $(\text{unit})$ and $(\text{other unit})$, change it into one of the forms above, and use it like this:
$$(\text{given value})(\text{unit})\times 1 = (\text{given value})(\text{unit})\times \frac{(\text{other number})(\text{other unit})}{(\text{number})(\text{unit})}$$
Since you have $(\text{unit})$ in both the numerator and denominator, you can cancel those out. Note that it's up to you to pick the right conversion factor to use so that you can cancel out units! Anyway, you'll be left with
$$\frac{(\text{given value})(\text{other number})}{(\text{number})}(\text{other unit})$$
Everything except for $(\text{other unit})$ is just a number, so now you can actually figure out the numerical value (e.g. plug things into your calculator if necessary).
If your given value has a compound unit (one made as a combination of other units, as in your examples), then you will need to use more than one unit conversion factor - basically, you will have to multiply by 1 more than once, using a different conversion factor each time. In your example, you would have to multiply by a conversion factor to go from $\mathrm{hm}$ to $\mathrm{km}$, and another one to go from $\mathrm{min.}$ to $\mathrm{hr.}$
Here's how that applies to your example, assuming Mark Eichenlaub is right about the mistake:
- $(\text{given value}) = 5.66$
For one part of the unit conversion:
- $(\text{unit}) = \mathrm{hm}$
- $(\text{other unit}) = \mathrm{km}$
- The conversion factor is $10\text{ hm} = 1\text{ km}$
For the other part of the unit conversion:
- $(\text{unit}) = \mathrm{min.}$
- $(\text{other unit}) = \mathrm{hr.}$
- The conversion factor is $60\text{ min.} = 1\text{ hr.}$
Note that the time units are in the denominator! But that doesn't really change anything - you still just need to arrange the conversion factor in the way that makes the original units cancel out.