Each vertex of this tree is either red or blue. How many possible trees are there? The question: Let $X$ denote the set of 'coloured trees' which result when each vertex of the tree is assigned one of the colours red or blue. How many different coloured trees of this kind are there?

I am not quite sure where to begin with this question. We are studying Burnside's Lemma and the orbit-stabilizer theorem, so I'm assuming these are necessary to solve the problem. 
What really throws me off is that there are only 7 vertices, so if each one is red or blue, do I consider all these cases?


*

*4 red, 3 blue

*3 red, 4 blue

*2 red, 5 blue

*5 red, 2 blue

*1 red, 6 blue

*6 red, 1 blue

*7 red, 0 blue

*0 red, 7 blue


The previous problem asked to find the automorphism group $G$ on the set of vertices of this tree, which I obtained. So maybe there is a way to apply $G$ to the trees with red or blue vertices in order to solve these? Any help is appreciated.

It sounds like a tree with $1,2,4$ blue is the same as a tree with $1,2,5$ blue.  I would count as follows:
First consider the subtree of $2,4,5$.  Vertex $2$ has two choices and vertices $4,5$ have three choices, so there are six ways to color the subtree.  Now for the whole tree, you have two choices for $1$.  You have $6$ ways to choose the colorings of the two subtrees so they match and $\frac 12\cdot 6 \cdot 5=15$ ways to choose two different colorings of the subtrees, for a total of $2(6+15)=42$
