(1) The coordinate of the symmetrical point b' of the point b relative to the straight line x=t is (2t+ 1, 0);

Because the distance from point B to x=t is t+ 1, the distance from b' to x=t is also t+ 1.

Then the coordinate is (2t+ 1, 0).

(2) The functional relationship between S and T needs to be solved in different situations:

The function formula of straight line AB obtained first, y=ax+b, is substituted into the coordinates of point A and point B, that is, there is

4a+b=0, b=2, that is, the function of straight line AB, y=- 1/2x+2,

Then the coordinates of point Q are: (t,-1/2t+2).

1, if 2t+ 1≥4, that is, 3/2 ≤ t < 4, then s is the area of △QPC.

If the length of PC is 4-t and the length of PQ is-1/2t+2, there is

s= 1/2(4-t)(- 1/2t+2)= 1/4t^2-2t+4

2. If 2t+ 1 < 4, that is, 0 < t < 3/2,

Let the graph of quadrilateral ABPQ symmetrical about the straight line x=t intersect with straight lines AB and OC at points M and N respectively.

Let the area of △QPC be S 1 and the area of △MNC be S2.

Then S=S 1-S2.

Because PC length is 4-t and PQ length is-1/2t+2, there is.

s 1= 1/2(4-t)(- 1/2t+2)= 1/4t^2-2t+4

Since point N is the symmetric point of point B about the line x=t, according to the result in (1),

If the coordinate of point N is (2t+ 1, 0), then NC=4-2t- 1=3-2t,

Because of the coordinates of point A', the symmetrical point of point A with respect to the straight line x=t is (2t, 2).

The function of the straight line A'N is: y=-2x+4t+2,

Since point M is the intersection of straight line A'N and straight line AB, the coordinate of point M is (8t/3,2-4t/3).

Then S2 =1/2 (3-2t) (2-4t/3) = 4t2/3-4t+3.

Then s = s1-S2 =1/4t2-2t+4-4t2/3+4t-3 =-13t2//kloc-0+2t+1.

A: When 3/2 ≤ t < 4, S = 1/4t 2-2t+4, and when 0 < t < 3/2, S =-13t2/12+2t/.