Not a complete answer, but working out the spin 2 massive bosons from spin 1 massive bosons is a fun little exercise. I find it confusing to use tables of Clebsch Gordan coefficients, and find it easier to just redo them from scratch.
The spin 1 rep has three vectors, $|1\rangle, |0\rangle, |-1\rangle$, which in the rest frame $k^\mu = (m,0,0,0)$ correspond to to the polarization vectors $\epsilon^\mu_1 = \tfrac{1}{\sqrt{2}}(0, 1, i, 0)$, $\epsilon^\mu_0 = (0, 0, 0, 1)$, $\epsilon^\mu_{-1} = \tfrac{1}{\sqrt{2}}(0, -1, i, 0)$. Note that they all satisfy the constraint $k^\mu \epsilon_\mu = 0$.
Now, under the spin $1$ rep, we have
\begin{align}
J_\pm^{(1)} &= J_x^{(1)} \pm i J_y^{(1)} \\
J_-^{(1)} | 1 \rangle &= \sqrt{2} |0 \rangle \\
J_-^{(1)} | 0 \rangle &= \sqrt{2} |-1 \rangle \\
J_z^{(1)} | m \rangle &= m |m\rangle
\end{align}
If we tensor two spin $1$ reps together, we get $1 \otimes 1 = 2 \oplus 1 \oplus 0$. We only want the $2$ rep. We can deduce the Clebsch Gordan decomposition for ourselves in the following manner. We begin by making a state with $m = 2$ and then using the lowering operator repeatedly on the state. So, in our tensor product $1 \otimes 1$, the Lie algebra operators becomes
$$
J_i \equiv 1 \otimes J_i^{(1)} + J_i^{(1)} \otimes 1.
$$
Now, the only option for the $m=2$ state is
$$
|2\rangle = |1 \rangle |1 \rangle
$$
and you can check that $J_z |2 \rangle = 2 |2 \rangle.$ Then we compute the action of the lowering operator using our expression for $J_\pm$, making sure to define our lowered states with the proper normalization so that the states we write as $|m\rangle$ all have norm $1$.
\begin{align}
J_-|2\rangle &= \sqrt{2}( |1\rangle |0 \rangle+|0\rangle |1 \rangle) \equiv 2 |1\rangle \\
J_-|1\rangle &= 2 |0\rangle |0\rangle + |1\rangle |-1\rangle + |-1\rangle |1\rangle \equiv \sqrt{6} | 0 \rangle \\
J_-|0\rangle &= \frac{1}{\sqrt{3}}(3 |-1\rangle |0\rangle + 3 |0\rangle |-1\rangle ) \equiv \sqrt{6} |-1\rangle\\
J_-|-1\rangle &= 2 |-1\rangle |-1 \rangle \equiv 2 |-1\rangle.
\end{align}
So, rephrasing our results,
\begin{align}
|2\rangle &= |1\rangle |1\rangle \\
|1\rangle &= \tfrac{1}{\sqrt{2}} (|1\rangle |0\rangle + |0\rangle |1\rangle )\\
|0\rangle &= \tfrac{1}{\sqrt{6}} ( 2 |0\rangle |0\rangle + |1\rangle |-1\rangle + |-1\rangle |1\rangle ) \\
|-1\rangle &= \tfrac{1}{\sqrt{2}} (|-1\rangle |0\rangle + |0\rangle |-1\rangle ) \\
|-2\rangle &= |-1\rangle |-1\rangle.
\end{align}
Now, to get our spin 2 polarization tensors $\epsilon^{\mu \nu}$, we just have to combine our spin 1 polarization vectors $\epsilon^\mu_{-1,0,+1}$ in the manner exactly above. So, to give one example,
$$
\epsilon^{\mu\nu}_0 = \frac{1}{\sqrt{6}}(2 \epsilon_0^\mu \epsilon_0^\nu + \epsilon_1^\mu \epsilon_{-1}^\nu+ \epsilon_{-1}^\mu \epsilon_1^\nu ).
$$
Notice that, by construction, these polarization tensors will satisfy $k_\mu \epsilon^{\mu \nu} = k_\nu \epsilon^{\mu \nu} = 0$. They also turn out to satisfy $\epsilon^\mu_{\; \mu} = 0$ and are symmetric. (Notice that we get a little taste of the slogan "gravity = E&M^2" from this method, even though we are working in the massive case.)
The results of taking these Kronecker products is
\begin{align}
\epsilon_2^{\mu \nu} &= \frac{1}{2} \begin{pmatrix} 0&0&0&0\\0&1&i&0\\0&i&-1&0\\0&0&0&0 \end{pmatrix} \\
\epsilon_1^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&0&0&1\\0&0&0&i\\0&1&i&0 \end{pmatrix} \\
\epsilon_0^{\mu \nu} &= \begin{pmatrix} 0&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&2 \end{pmatrix} \\
\epsilon_{-1}^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&0&0&-1\\0&0&0&i\\0&-1&i&0 \end{pmatrix} \\
\epsilon_{-2}^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&1&-i&0\\0&-i&-1&0\\0&0&0&0 \end{pmatrix} \\
\end{align}
So, hopefully you can see that Clebsch Gordan coefficients are not so scary, and that you are always free to rederive them for yourself. Admittedly, we did not work from tensoring together four copies of the spin $1/2$ case, but rather used two copies of the spin $1$ case to ease our burden, but the principle is what is important.
