Let f(x) = cases (1)/(3), & x ≤ π/2 (b(1- x))/((π-2x)^2), & x > π/2...

Since $f(x)$ is continuous at $x = \pi/2$, the right-hand limit must equal the value of the function at $x = \pi/2$.

$\lim_{x \to \pi/2^+} f(x) = f(\pi/2)$

$\lim_{x \to \pi/2^+} \dfrac{b(1-\sin x)}{(\pi-2x)^2} = \dfrac{1}{3}$

Let $x = \pi/2 + h$. As $x \to \pi/2^+$, $h \to 0^+$.

$\lim_{h \to 0^+} \dfrac{b(1-\sin(\pi/2+h))}{(\pi-2(\pi/2+h))^2} = \dfrac{1}{3}$

$\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{(-2h)^2} = \dfrac{1}{3}$

$\lim_{h \to 0^+} \dfrac{b(1-\cos h)}{4h^2} = \dfrac{1}{3}$

Using the standard limit $\lim_{h \to 0} \dfrac{1-\cos h}{h^2} = \dfrac{1}{2}$, we get:

$\dfrac{b}{4} \times \dfrac{1}{2} = \dfrac{1}{3} \Rightarrow \dfrac{b}{8} = \dfrac{1}{3} \Rightarrow b = \dfrac{8}{3}$

The upper limit of the integral is $3b - 6 = 3\left(\dfrac{8}{3}\right) - 6 = 2$.

The integral to evaluate is $I = \displaystyle\int_{0}^{2} |x^2+2x-3|\,dx$.

Factoring the quadratic expression gives $x^2+2x-3 = (x+3)(x-1)$.

For $x \in [0, 1]$, $(x+3)(x-1) \leq 0$, so $|x^2+2x-3| = 3 - 2x - x^2$.

For $x \in [1, 2]$, $(x+3)(x-1) \geq 0$, so $|x^2+2x-3| = x^2+2x-3$.

Splitting the integral at $x = 1$:

$I = \displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx + \displaystyle\int_{1}^{2} (x^2+2x-3)\,dx$

Evaluating the first integral:

$\displaystyle\int_{0}^{1} (3 - 2x - x^2)\,dx = \left[ 3x - x^2 - \dfrac{x^3}{3} \right]_0^1 = 3 - 1 - \dfrac{1}{3} = \dfrac{5}{3}$

Evaluating the second integral:

$\displaystyle\int_{1}^{2} (x^2+2x-3)\,dx = \left[ \dfrac{x^3}{3} + x^2 - 3x \right]_1^2$

$= \left( \dfrac{8}{3} + 4 - 6 \right) - \left( \dfrac{1}{3} + 1 - 3 \right) = \left( \dfrac{8}{3} - 2 \right) - \left( \dfrac{1}{3} - 2 \right) = \dfrac{7}{3}$

Adding the two parts:

$I = \dfrac{5}{3} + \dfrac{7}{3} = \dfrac{12}{3} = 4$

Answer: $4$