The sum of all local minimum values of the function
$$f(x) = \left\{\begin{array}{l l}1-2x, & \quad {x<-1}\\ \frac{1}{3}(7+2|x|), & \quad {-1\leq x\leq 2}\\\frac{11}{18}(x-4)(x-5), & \quad {x>2}\\ \end{array}\right.$$ is

$$x < -1$$: $$f(x) = 1-2x$$ (decreasing).

 $$-1 \le x \le 2$$: $$f(x) = \frac{1}{3}(7+2|x|)$$.

 At $$x=0$$, there is a local minimum because the absolute value $$|x|$$ turns from decreasing to increasing.

$$f(0) = 7/3$$.

$$x > 2$$: $$f(x) = \frac{11}{18}(x-4)(x-5)$$.

 This is a parabola opening upward. Vertex at $$x = 4.5$$.

 $$f(4.5) = \frac{11}{18}(0.5)(-0.5) = \frac{11}{18}(-\frac{1}{4}) = -\frac{11}{72}$$.

 Check boundaries:

 At $$x = -1$$: $$f(-1^-) = 3$$, $$f(-1^+) = 3$$ continuous.

 At $$x = 2$$: $$f(2^-) = 11/3 = 88/24$$, $$f(2^+) = \frac{11}{18}(-2)(-3) = \frac{66}{18} = 11/3$$. Continuous.

 Sum of minima: $$f(0) + f(4.5) = \frac{7}{3} - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}$$.

Answer: A ($$\frac{157}{72}$$)