Answer
(a) The rate of deceleration is $180 ~m/s^2$
(b) Loosening the net would make the net safer by decreasing the rate of deceleration.
Work Step by Step
(a) First we need to find the velocity when the person arrives at the net.
$v^2 = v_0^2 + 2a(\Delta y)$
$v = \sqrt{v_0^2 + 2a(\Delta y)} = \sqrt{0 + (2)(9.80 ~m/s^2)(18.0 ~m)}$
$v = 18.8 ~m/s$
We can use this velocity as $v_0$ in the next part of the problem to find the rate of deceleration.
$a = \frac{v^2 - v_0^2}{2 \Delta y} = \frac{0 - (18.8 ~m/s)^2}{(2)(1.0 ~m)} = -180 ~m/s^2$
The average rate of deceleration is $180 ~m/s^2$
(b) If we loosen the net, the stopping distance $\Delta y$ would increase. From the equation above, we can see that increasing $\Delta y$ would decrease the rate of deceleration. Therefore, loosening the net would make the net safer by decreasing the rate of deceleration (as long as the person stops before hitting the ground!).
