5. Place a sphere inside a cone, which is tangent to the cone's side and base. Then the maximum ratio of the sphere's surface area to the cone's surface area is $\qquad$

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5. $\frac{1}{2}$.

Let the radius of the base of the cone be $r$, and the angle between the slant height and the base be $2 \theta$. Then the radius of the inscribed sphere $R=r \tan \theta$.
$$
\begin{array}{l}
\text { By } S_{\text {sphere }}=4 \pi R^{2}=4 \pi r^{2} \tan ^{2} \theta, \\
S_{\text {cone }}=\pi r^{2}\left(1+\frac{1}{\cos 2 \theta}\right), \\
\text { then } \frac{S_{\text {sphere }}}{S_{\text {cone }}}=\frac{4 \tan ^{2} \theta}{1+\frac{1}{\cos 2 \theta}}=\frac{2 \sin ^{2} \theta \cdot \cos 2 \theta}{\cos ^{4} \theta} \\
=-4+\frac{6}{\cos ^{2} \theta}-\frac{2}{\cos ^{4} \theta} \\
=-2\left(\frac{1}{\cos ^{2} \theta}-\frac{3}{2}\right)^{2}+\frac{1}{2} \leqslant \frac{1}{2} .
\end{array}
$$

Therefore, when $\cos \theta=\sqrt{\frac{2}{3}}$, i.e., $\cos 2 \theta=\frac{1}{3}$, the equality holds.
Thus, the maximum value is $\frac{1}{2}$.