\section*{11. ACD}
Solution: The radius of the circle is $r = 4$. The equation of line $AB$ is $y = - \frac { 1 } { 2 } x + 2$. Drawing a line through $P$ parallel to $AB$, the distance between the two lines is $d = \frac { \left| \frac { 15 } { 2 } - 2 \right| } { \sqrt { 1 + \left( - \frac { 1 } { 2 } \right) ^ { 2 } } } = \frac { 11 } { \sqrt { 5 } }$. Since $d < 6$, the maximum distance from $P$ to line $AB$ is $d + r$, so A is correct; the minimum distance from $P$ to line $AB$ is $d - r < 2$, so B is incorrect; the extremum of $\angle P A B$ is attained when $PB$ is tangent to the circle. We have $O B = \sqrt { 5 ^ { 2 } + 3 ^ { 2 } } = \sqrt { 34 }$. Since $PB$ is tangent to $OB$, we have $PB \perp O B$. By the Pythagorean theorem, $| P B | = \sqrt { 34 - 4 ^ { 2 } } = \sqrt { 18 }$. The two tangent lines from a point to a circle have equal length, so C and D are correct. The answer is $ACD$.

When $A _ { 1 } P \perp B P$, there are two points $P$ satisfying the condition, so C is incorrect. For option D, let $E$ be the midpoint of $C C _ { 1 }$, and let $G$ be the center of rectangle $A A _ { 1 } B _ { 1 } B$. When $\mu = \frac { 1 } { 2 }$, $P$ is a point on $EF$. Since $E G \perp$ plane $A A _ { 1 } B _ { 1 }$ and $A _ { 1 } B \perp A B _ { 1 }$, we have $A _ { 1 } B \perp$ plane $E A B _ { 1 }$. When $P$ coincides with $E$, the condition is satisfied. There is a unique plane perpendicular to line $A _ { 1 } B$ passing through such points.

\section*{III. Fill in the Blank Questions}
13. 1

Solution: Setting $f ( x ) = f ( - x )$ gives $x ^ { 3 } \left( 2 ^ { x } + 2 ^ { - x } \right) ( a - 1 ) = 0$ for all $x$, so $a = 1$.