= Solution
The <Clebsch-Gordan decomposition for SU2> is
$$
\boxed{V_{j_1}\otimes V_{j_2}\cong\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}V_j,\qquad \dim V_j=2j+1,}
$$
where $j$ increases by one and every summand occurs once. The <tensor product of Lie algebra representations> is antihermitian for the product inner product, so an invariant subspace has an invariant orthogonal complement; this gives complete reducibility. To see the particular summands, suppose $j_1\ge j_2$. The product <weight diagram> has one state of weight $j_1+j_2$, two of weight $j_1+j_2-1$, and increasing multiplicities down to the plateau determined by $j_1-j_2$. Subtracting the weight string of $V_{j_1+j_2}$ leaves a single highest state of weight $j_1+j_2-1$. Repeating <highest-weight character subtraction> yields exactly the displayed sequence, down to $j_1-j_2$. Its dimensions exhaust the tensor product because
$$
\sum_{j=j_1-j_2}^{j_1+j_2}(2j+1)=(2j_1+1)(2j_2+1).
$$
The case $j_2\ge j_1$ is identical with the factors exchanged.
Let $C^j_m(m_1,m_2)=\langle m_1:m_2|j,m\rangle$ be the <Clebsch-Gordan coefficients>. Since the total $J_3$ is $J_3^{(1)}\otimes1+1\otimes J_3^{(2)}$, orthogonality of its <eigenspaces> gives the selection rule \b[$C^j_m(m_1,m_2)=0$ unless $m_1+m_2=m$]. Normalize the coupled states by lowering from a unit <highest-weight vector>, as in part (i); a common phase for an entire summand is immaterial. Then
$$
J_\mp|j,m\rangle=\sqrt{\frac{(j\pm m)(j\mp m+1)}2}|j,m\mp1\rangle.
$$
Take the inner product with $\langle m_1:m_2|$. On the other hand, the total <ladder operator> is $J_\mp^{(1)}\otimes1+1\otimes J_\mp^{(2)}$, and $J_\mp^\dagger=J_\pm$. Its action on the product bra is therefore
$$
\begin{aligned}
\langle m_1:m_2|J_\mp
&=\sqrt{\frac{(j_1\mp m_1)(j_1\pm m_1+1)}2}\langle m_1\pm1:m_2|\\
&\quad+\sqrt{\frac{(j_2\mp m_2)(j_2\pm m_2+1)}2}\langle m_1:m_2\pm1|.
\end{aligned}
$$
Equating the two evaluations and multiplying by $\sqrt2$ proves both signs of the <ladder recurrence for Clebsch-Gordan coefficients>:
$$
\boxed{\begin{aligned}
\sqrt{(j\pm m)(j\mp m+1)}\,C^j_{m\mp1}(m_1,m_2)
&=\sqrt{(j_1\mp m_1)(j_1\pm m_1+1)}\,C^j_m(m_1\pm1,m_2)\\
&\quad+\sqrt{(j_2\mp m_2)(j_2\pm m_2+1)}\,C^j_m(m_1,m_2\pm1).
\end{aligned}}
$$
Terms with an index outside its allowed weight range are zero. Merely specifying eigenvectors of $J_3$ would allow unrelated phases for different $m$; the coherent positive-ladder convention above is what makes these recurrences hold with the displayed positive square roots.
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