1)What does it mean for something to have a spectral decomposition?
Since you have your eigenvectors, how can you show that it has one?
2)You should be able to find the projections easily enough
3)I'm not quite sure what 'use the spectral theorem' means since there is no 'spectral theorem' on it's own. I assume it means just work out the decomposition and show that it agrees with A.

The Q is changing the basis you are working in.
If [itex]a[/itex], [itex]b[/itex] are the eigenvectors of [itex]T[/itex] and [itex]\lambda_1[/itex], [itex]\lambda_2[/itex] the eigenvalues we have

If we use the basis where the vector (1,0) ~ a and (0,1) ~ b then T is obviously diagonal. What the Q does is takes the vectors a and b and turns them into (1,0) and (0,1) respectively.

If we have a as the first row and b as the second row in Q then [itex]Q.(1,0) = a[/itex] and [itex]Q.(0,1) = b[/itex]. So then [itex]Q^{-1}.a = (1,0)[/itex] and [itex]Q^{-1}.a = (0,1)[/itex]

What your [itex]T=Q\ D\ Q^{-1}[/itex] decomposition is doing is when it acts on vector x is this;
1. Change the basis of x from (1,0) and (0,1) to a and b
2. Apply the diagonal matrix to a and b, a simple multiplication [itex]a \rightarrow \lambda_1 a[/itex], [itex]b \rightarrow \lambda_2 b[/itex]
3. Change the basis back to (1,0) and (0,1) from a and b