16. C;\\
17. C;\\
18. D.

\section*{III. Solution Problems}
19. The original expression $= \lg ( \sin x + \cos x ) + \lg ( \cos x + \sin x ) - \lg ( \sin x + \cos x ) ^ { 2 } = 0$ .\\
20. (1) When $n = 1$ , $a _ { 1 } = - 14$ ; when $n \geq 2$ , $a _ { n } = S _ { n } - S _ { n - 1 } = - 5 a _ { n } + 5 a _ { n - 1 } + 1$ , so $a _ { n } - 1 = \frac { 5 } { 6 } \left( a _ { n - 1 } - 1 \right)$ ,\\
Since $a _ { 1 } - 1 = - 15 \neq 0$ , the sequence $\left\{ a _ { n } - 1 \right\}$ is a geometric sequence;\\
(2) From (1): $a _ { n } - 1 = - 15 \cdot \left( \frac { 5 } { 6 } \right) ^ { n - 1 }$ , we get $a _ { n } = 1 - 15 \cdot \left( \frac { 5 } { 6 } \right) ^ { n - 1 }$ , thus\\
$S _ { n } = 75 \cdot \left( \frac { 5 } { 6 } \right) ^ { n - 1 } + n - 90 \left( n \in \mathbf { N } ^ { * } \right) ;$\\
Solving the inequality $S _ { n } < S _ { n + 1 }$ , we get $\left( \frac { 5 } { 6 } \right) ^ { n - 1 } < \frac { 2 } { 5 } , n > \log _ { \frac { 5 } { 6 } } \frac { 2 } { 25 } + 1 \approx 14.9$ , when $n \geq 15$ , the sequence $\left\{ S _ { n } \right\}$\\
is monotonically increasing;\\
Similarly, when $n \leq 15$ , the sequence $\left\{ S _ { n } \right\}$ is monotonically decreasing; therefore when $n = 15$ , $S _ { n }$ attains its minimum value.\\
21. (1) Let the slant height of the cylindrical lantern be $l$ , then $l = 1.2 - 2 r ( 0 < r < 0.6 ) , S = - 3 \pi ( r - 0.4 ) ^ { 2 } + 0.48 \pi$ , so when $r = 0.4$ , $S$ attains its maximum value of approximately $1.51$ square meters;\\
(2) When $r = 0.3$ , $l = 0.6$ , establishing a spatial rectangular coordinate system, we obtain $\overrightarrow { A _ { 1 } B _ { 3 } } = ( - 0.3,0.3,0.6 )$ , $\overrightarrow { A _ { 3 } B _ { 5 } } = ( - 0.3 , - 0.3,0.6 )$ ,

Let the angle between vectors $\overrightarrow { A _ { 1 } B _ { 3 } }$ and $\overrightarrow { A _ { 3 } B _ { 5 } }$ be $\theta$ , then $\cos \theta = \frac { \overrightarrow { A _ { 1 } B _ { 3 } } \cdot \overrightarrow { A _ { 3 } B _ { 5 } } } { \left| \overrightarrow { A _ { 1 } B _ { 3 } } \right| \cdot \left| \overrightarrow { A _ { 3 } B _ { 5 } } \right| } = \frac { 2 } { 3 }$ ,\\
therefore the angle formed by the skew lines containing $A _ { 1 } B _ { 3 }$ and $A _ { 3 } B _ { 5 }$ is $\arccos \frac { 2 } { 3 }$ .\\
22. (1) $x \in ( - \infty , - \sqrt { 2 } ) \cup ( \sqrt { 2 } , + \infty )$ ;\\
(2) For any two distinct positive numbers $a$ and $b$ , we have $a ^ { 3 } + b ^ { 3 } > 2 a b \sqrt { a b } , ~ a ^ { 2 } b + a b ^ { 2 } > 2 a b \sqrt { a b }$ . Since $\left| a ^ { 3 } + b ^ { 3 } - 2 a b \sqrt { a b } \right| - \left| a ^ { 2 } b + a b ^ { 2 } - 2 a b \sqrt { a b } \right| = ( a + b ) ( a - b ) ^ { 2 } > 0$ , we have $\left| a ^ { 3 } + b ^ { 3 } - 2 a b \sqrt { a b } \right| > \left| a ^ { 2 } b + a b ^ { 2 } - 2 a b \sqrt { a b } \right|$ , that is, $a ^ { 3 } + b ^ { 3 }$ is farther from $2 a b \sqrt { a b }$ than $a ^ { 2 } b + a b ^ { 2 }$ ;\\
(3) $f ( x ) = \left\{ \begin{array} { l l } | \sin x | , & x \in \left( k \pi + \frac { \pi } { 4 } , k \pi + \frac { 3 \pi } { 4 } \right) \\ | \cos x | , & x \in \left( k \pi - \frac { \pi } { 4 } , k \pi + \frac { \pi } { 4 } \right) \end{array} \right.$ ,\\
Properties: $1 ^ { \circ }$ $f ( x )$ is an even function, its graph is symmetric about the $y$-axis; $2 ^ { \circ }$ $f ( x )$ is a periodic function with minimum positive period $T = \frac { \pi } { 2 }$ ;\\
$3 ^ { \circ }$ The function $f ( x )$ is monotonically increasing on the interval $\left( \frac { k \pi } { 2 } - \frac { \pi } { 4 } , \frac { k \pi } { 2 } \right]$ and monotonically decreasing on the interval $\left[ \frac { k \pi } { 2 } , \frac { k \pi } { 2 } + \frac { \pi } { 4 } \right)$ , $k \in \mathbf { Z }$ ; $4 ^ { \circ }$ The range of the function $f ( x )$ is $\left( \frac { \sqrt { 2 } } { 2 } , 1 \right]$ .\\
23. (1) $M \left( \frac { a } { 2 } , - \frac { b } { 2 } \right)$ ;\\
(2) From the system of equations $\left\{ \begin{array} { l } y = k _ { 1 } x + p \\ \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 \end{array} \right.$ , eliminating $y$ gives the equation $\left( a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } \right) x ^ { 2 } + 2 a ^ { 2 } k _ { 1 } p x + a ^ { 2 } \left( p ^ { 2 } - b ^ { 2 } \right) = 0$ ,\\
Since the line $l _ { 1 } : y = k _ { 1 } x + p$ intersects the ellipse $\Gamma$ at points $C$ and $D$ ,\\
we have $\Delta > 0$ , that is, $a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } - p ^ { 2 } > 0$ ,\\
Let $C \left( x _ { 1 } , y _ { 1 } \right)$ , $D \left( x _ { 2 } , y _ { 2 } \right)$ , and the midpoint of $C D$ be $\left( x _ { 0 } , y _ { 0 } \right)$ ,\\
then $\left\{ \begin{array} { l } x _ { 0 } = \frac { x _ { 1 } + x _ { 2 } } { 2 } = - \frac { a ^ { 2 } k _ { 1 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } \\ y _ { 0 } = k _ { 1 } x _ { 0 } + p = \frac { b ^ { 2 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } \end{array} \right.$ ,\\
From the system of equations $\left\{ \begin{array} { l } y = k _ { 1 } x + p \\ y = k _ { 2 } x \end{array} \right.$ , eliminating $y$ gives the equation $\left( k _ { 2 } - k _ { 1 } \right) x = p$ ,\\
Since $k _ { 2 } = - \frac { b ^ { 2 } } { a ^ { 2 } k _ { 1 } }$ , we have $\left\{ \begin{array} { l } x = \frac { p } { k _ { 2 } - k _ { 1 } } = - \frac { a ^ { 2 } k _ { 1 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } = x _ { 0 } \\ y = k _ { 2 } x = \frac { b ^ { 2 } p } { a ^ { 2 } k _ { 1 } ^ { 2 } + b ^ { 2 } } = y _ { 0 } \end{array} \right.$ ,\\
Therefore $E$ is the midpoint of $C D$ ;\\
(3) Steps for constructing points $P _ { 1 }$ and $P _ { 2 }$ : $1 ^ { \circ }$ Find the midpoint $E \left( - \frac { a ( 1 - \cos \theta ) } { 2 } , \frac { b ( 1 + \sin \theta ) } { 2 } \right)$ of $P Q$ , $2 ^ { \circ }$ Find the slope of line $O E$ as $k _ { 2 } = - \frac { b ( 1 + \sin \theta ) } { a ( 1 - \cos \theta ) }$ ,\\
$3 ^ { \circ }$ From $\overrightarrow { P P _ { 1 } } + \overrightarrow { P