$3 \sin \theta+5 \cos \theta=5 \Rightarrow 3 \sin \theta=5(1-\cos \theta)$ $\Rightarrow 3.2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=5.2 \sin ^{2} \frac{\theta}{2}$ $\left(\because \sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)$ and $\left.1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}\right)$
$$
\begin{array}{l}
\Rightarrow \tan \frac{\theta}{2}=\frac{3}{5} \\
\text { Now, } 5 \sin \theta-3 \cos \theta \\
=5 \cdot \frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}-3 \cdot \frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}} \\
=5 \cdot \frac{2 \cdot \frac{3}{5}}{\left(1+\frac{9}{25}\right)}-3 \cdot \frac{\left(1-\frac{9}{25}\right)}{\left(1+\frac{9}{25}\right)} \\
=\frac{6-3 \cdot \frac{16}{25}}{1+\frac{9}{25}}=\frac{150-48}{34}=\frac{102}{34}=3
\end{array}
$$