Questions about rigid rotating dumbbell molecule

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I would like some help with the following problems

1. Consider in R^{j}={f : f = \Sigma^{l}_{m=-l} a^{m}f^{l}_{m}} the operator \stackrel{\rightarrow}{e}\bullet\stackrel{\rightarrow}{J}, where \stackrel{\rightarrow}{e} is a unit vector in 3-dimensional space.
(a) Calculate the probabilities for all eigenvalues of \stackrel{\rightarrow}{e}\bullet\stackrel{\rightarrow}{J} in the state W^{j} = Tr(\Lambda^{j})^{-1}\Lambda^{j}, where \Lambda^{j} is the projection operator onto Rj .
(b) Calculate the expectation value for the component J_{2} in the state W^{j} .

2. What spaces R^{l'}_{m'} are obtained when the operators (Q_{\stackrel{+}{-}})^{2} act on the space R^{l}_{m}?

3. Consider the rigidly rotating dumbbell molecule and let Q_{i}, J_{i}, i = 1, 2, 3 denote the position and angular momentum operators.
(a) Find a complete system of commuting observables.
(b) Explain the physical meaning of these observables and explain the meaning of their eigenvalues.
(c) Prove that the operators of your choice form a system of commuting observables.Number 1 is really confusing me since we need the probabilities for ALL eigenvalues of \stackrel{\rightarrow}{e}\bullet\stackrel{\rightarrow}{J}, and we don't know what 'j' is.
To find, say, the probabilities for the eigenvalues of J_{3}, is it just

\Sigma^{r}_{s=-r}\Sigma^{l}_{m=-l} |<a^{s}f^{r}_{s} | J_{3} | a^{m}f^{l}_{m}>| ^{2} = \Sigma^{l}_{m=-l}m^{2} ?

I am clueless as to how to solve #2

For #3, I found that because [J_{i}, Q_{j}] = i*h*\epsilon_{i,j,k}*Q_{k}, then they don't commute. Thus the CSCO is {Q_{I}, Q_{j}, Q_{k}}. Is this right?
 
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For part (b), I think the physical meaning of these observables is the position and angular momentum of the rigidly rotating dumbbell molecule. The eigenvalues would correspond to its position and angular momentum respectively. For part (c), since [Q_{i}, Q_{j}] = 0, [Q_{i}, Q_{k}] = 0, and [Q_{j}, Q_{k}] = 0, then it is proven that they form a system of commuting observables. I hope this helps.
 
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