Eigenvalue/Eigenstate of Hamilton with 2 Spin particles I got a task, I don't quite know how to solve.

I've got the following Hamiltonian: 
  $$
\hat H = \frac{B}{\hbar^2}\hat{\mathbf S}_1\cdot \hat{\mathbf S}_2+\frac{C}{\hbar}\left(\hat S_{1z}+\hat S_{2z}\right), \qquad \hat{\mathbf S}_{j=1,2}=(\hat S_{jx},\hat S_{jy},\hat S_{jz}).
$$
  ($B,c$ constants)
My task is to calculate the Eigenvalues and Eigenstates of $\hat{H}$ for:
  
  
*
  
*Two spin 1/2 particles
  
*One spin 1/2 and one spin 1 particle
I got a Tip. I have to write Hamiltonian with the following operators $\hat{S}^2,\hat{S}_z,\hat{S}_1^2,\hat{S}_2^2$, where $\boldsymbol{\hat{S}}=\boldsymbol{\hat{S}_1}+\boldsymbol{\hat{S}_2}$

This was no Problem:
$\hat{H}=\frac{B}{\hbar^2}(\hat{S}^2-\hat{S}_1^2-\hat{S}_2^2)+\frac{c}{\hbar}\hat{S}_z$
Now I'm pretty much stuck. My idea was to give out these operators as matrix and the rest is simple linear algebra. But I quit don't understand that. Furthermore I heard that I need the Clebsch-Gordan coefficients, but don't exactly know where.
 A: Notice that $S_j = S^1_j\otimes 1_2 + 1_1\otimes S^2_j$, with $S_j$ living in $\mathcal{H_1}\otimes\mathcal{H_2}$.
The Hamiltonian commutes with $(S^2, S_z)$, therefore it can be diagonalised onto their eigenstates; as such, the collection $|S\, M_S\rangle$ (with corresponding angular momentum eigenvalues) spans the set of possible eigenstates of $H$. Using the initial definition of $S$ in terms of $S_1,S_2$ one can turn each $|S\, M_S\rangle$ into an expansion in terms of $|s_1\, m_{s_1}\rangle$ and $|s_2\, m_{s_2}\rangle$ with some more or less complicated coefficients (Clebsch-Gordan).
A: *

*The space of two spin $\frac{1}{2}$ is spanned by the basis $|0,0>, |1,1>, |1,0>, |1,-1>$ (written in $|S,S_z>$ form). Now write down the Hamiltonian in this basis in matrix form and find eigenfunction and eigenvalues of the corresponding matrix. In this case the matrix turns out to be diagonal.
$$
  H=
  \left[ {\begin{array}{cccc}
   -\frac{3B}{2} & 0 & 0 & 0 \\
   0 & \frac{B}{2}+C & 0 & 0 \\
   0 & 0 & \frac{B}{2} & 0 \\
   0 & 0 & 0 & \frac{B}{2}-C\\
  \end{array} } \right]
$$
So, $|0,0>, |1,1>, |1,0>, |1,-1>$ are the eigenstates with eigenvalue $-\frac{3B}{2}$, $\frac{B}{2}+C$, $\frac{B}{2}$, $\frac{B}{2}-C$ respectively. You will need Clebsh Gordon coefficient if you want to write the states in $|S_{1z},S_{2z}>$ basis.

*Follow the same procedure. In this case the basis will be $|\frac{3}{2},\pm\frac{3}{2}>$, $|\frac{3}{2},\pm\frac{1}{2}>$, $|\frac{1}{2},\pm\frac{1}{2}>$
