What is this circuits output and how can I simplify it?
What you should do is write down a truth table. For every state in the truth table, analyze every gate and work your way to the output. You can include this in the truth table - give the gates a name, and their own collum. 

A B C | Ga1 Ga2 Ga3 | OUT
0 0 0 |  1   0   1  |  0
0 0 1 |  0   1   0  |  1
0 1 0 | ...
0 1 1 | ...
1 0 0 |
1 0 1 | 
1 1 0 |
1 1 1 |


EDIT: This is perhaps not the best way to do it here.
We can view every gate as a equation - the basic building blocks have quite simple equations: AND multiplies, OR Adds.

In this case, let's name every intermediate node. We can then look at the equation from an intermediate node to it's inputs. This will end up giving us an equation we can then simplify.

I prefer not to solve your specific example - let me build another so we can study this, and you can then apply this method to the exam question. Take the following:


I named the intermediate nodes \$Y_{1}\$ and \$Y_{2}\$.

Now we can see that:

$$ Y_{1} = A \cdot B $$
$$ Y_{2} = B + C $$

and that the output \$Y\$
$$Y = Y_{1} \cdot Y_{2} $$
$$ \Leftrightarrow Y = (A \cdot B) \cdot (B + C) $$

Now you can use boolean algebra to simplify this and see if you can come to a simpler equation, perhaps provided some constraints. Let me give it a try and see how we can build our circuit with less gates, using only 2-input gates:

$$Y = (A \cdot B) \cdot (B + C) = A \cdot B \cdot B + A \cdot B \cdot C = A \cdot B$$

Obvioulsy this was a very simple example, but we see that we can build the same logic diagram by using just one gate!



I hope this is enought to get you started!

Thanks Trevor for pointing out a (silly) mistake on my part in the first version.