As shown in the figure: it is known that the midpoint a (a, b) point B (a, 0) in the plane rectangular coordinate system satisfies | 2a-b | + (B-4) 2 = 0 (1) (2) given point C (0, b), point P starts from point B and moves at a speed of 1 unit per second in the negative direction of x-axis. At the same time, point Q starts from point C and moves at a speed of 2 units per second in the negative direction of y-axis. At a certain time, as shown in the figure, and s yin = 12s quadrilateral ocab, the moving time of point P is calculated? (3) Under the condition of (2), AQ intersects the x-axis at m, makes the angle bisector of ∠ ACO and ∠ AMB intersects at point n, and judges whether ∠ n - ∠ APB - ∠ PAQ ∠ AQC is a fixed value. If it is a fixed value, calculate the value; if not, explain the reason

As shown in the figure: it is known that the midpoint a (a, b) point B (a, 0) in the plane rectangular coordinate system satisfies | 2a-b | + (B-4) 2 = 0 (1) (2) given point C (0, b), point P starts from point B and moves at a speed of 1 unit per second in the negative direction of x-axis. At the same time, point Q starts from point C and moves at a speed of 2 units per second in the negative direction of y-axis. At a certain time, as shown in the figure, and s yin = 12s quadrilateral ocab, the moving time of point P is calculated? (3) Under the condition of (2), AQ intersects the x-axis at m, makes the angle bisector of ∠ ACO and ∠ AMB intersects at point n, and judges whether ∠ n - ∠ APB - ∠ PAQ ∠ AQC is a fixed value. If it is a fixed value, calculate the value; if not, explain the reason

(1) ∵| - 2a-b | + (B-4) 2 = 0. ∵ 2a-b = 0, B-4 = 0, ∵ a = 2, B = 4, ∵ the coordinates of point a are (2,4), and the coordinates of point B are (2,0); (2) as shown in Figure 2, if the motion time of point P is ts, then T & gt; 2, so the coordinates of point P are (2-T, 0), and the coordinates of point q are (0, 4-2t). Let the analytic formula of the line AQ be y = KX + 4-2t, and substitute a (2, 4) into 2K + 4-2t = 4, and the solution is k = T-1. The analytic formula of the line AQ is y = (t-1) x + 4-2t, and the coordinates of the intersection of the line AQ and the X axis are (2t-4t-1, 0). S shadow = 12 (2t-4t-1 + T-2) × 4 + 12 × 2t-4t-1 × (2t-4), and s yin = 12s quadrilateral ocab, 12 (2) 4t-4t-4t-4t-1 + T-2) × 4 + 12 × 2t-4t-4t-1 × (2t-4) = 12 × 2 × 4, finishing 2t2-7t + 4 = 0, solving T1 = 7 + 174, T2 = 7-174 (rounding off), the time of point P moving is 7 + 174s; (3) n - APB - PAQ, AQC is the fixed value. The reasons are as follows: as shown in Figure 3, 2t2-7t-7t-7t + 4 = 0, 2t2-7t-7t + 4 = 0, the solution is T1 = 7 + 7 + 174, T1 = 7 + 174, T1 = 7 + 174, T2 = 7 + 174, T2 = 7 = 7 + 174 (rounded off), the time of point P is 7 + 7 + 174s; the time of point P is 7 + 7 + 174s; the time of point P is 7 + 174s; 3; (3) n; (3) n is the fixed value how to use it N + ∠ 1, 〈 45 ° + 2 ∠ 1 = ∠ n + ∠ 1, 〉 n = 45 ° + ∠ 1, ∫ AMB = ∠ APB + ∠ PAQ, ∫ APB + ∠ PAQ = 2 ∠ 1, ∫ AQC + ∠ omq = 90 ° and ∠ omq = 2 ∠ 1, ∫ AQC = 90 ° - 2 ∠ 1, ∫ n - ∠ APB - ∠ PAQ ∠ AQC = 45 ° + ∠ 1-2 ∠ 1 = 12