22% ** Let $0<\theta<\pi$. Then the maximum value of $\sin \frac{\theta}{2}(1+\cos \theta)$ is
First, transform the different angles in the original expression to the same angle: $\sin \frac{\theta}{2}(1+\cos \theta)=2 \sin \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}$. Directly finding its maximum value is difficult. Noting the square sum relationship of sine and cosine functions, square both sides of the original expression: $4 \sin ^{2} \frac{\theta}{2} \cos ^{4} \frac{\theta}{2}=2 \cdot 2 \sin ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2} \leqslant 2 \cdot\left(\frac{2 \sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}}{3}\right)^{3}$ $=\frac{16}{27}$. Since $0 < \theta < \pi$, $\sin \frac{\theta}{2}(1+\cos \theta) > 0$, so $\sin \frac{\theta}{2}(1+\cos \theta) \leqslant \frac{4 \sqrt{3}}{9}$, with equality holding if and only if $2 \sin ^{2} \frac{\theta}{2}=\cos ^{2} \frac{\theta}{2}$, i.e., $\theta=2 \operatorname{arccot} \sqrt{2}$. Therefore, the maximum value sought is $\frac{4 \sqrt{3}}{9}$.