7. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then
$$
\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=
$$
ニ、7.9.
From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta$, then
$$
\begin{array}{l}
(\cos \theta-1)^{2}=4 \Rightarrow \cos \theta=-1 . \\
\text { Therefore, }\left(4+\cos ^{3} \theta\right)\left(3+\sin ^{3} \theta\right)=9 .
\end{array}
$$