= Solution
The <Poisson bracket> is $\{f,g\}=\sum_a(f_{q_a}g_{p_a}-f_{p_a}g_{q_a})$. For a function with no explicit time dependence, the chain rule and <Hamilton equations> give
$$
\frac{df}{dt}=\sum_a(f_{q_a}\dot q_a+f_{p_a}\dot p_a)=\sum_a(f_{q_a}H_{p_a}-f_{p_a}H_{q_a})=\boxed{\{f,H\}}.
$$
With $L_b=\epsilon_{bcd}x_cp_d$, direct differentiation gives
$$
\boxed{\{p_a,L_b\}=-\epsilon_{bad}p_d=\epsilon_{abd}p_d,\qquad \{x_a,L_b\}=\epsilon_{abc}x_c.}
$$
Apply the product rule to $L_a=\epsilon_{acd}x_cp_d$ and substitute these two brackets. Contracting the epsilon symbols yields $\boxed{\{L_a,L_b\}=\epsilon_{abc}L_c}$. Equivalently, the two product terms are precisely the infinitesimal rotation of the vector $x\times p$ about the $b$ axis, with the same bracket convention as the position and momentum vectors.
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