To get the best estimate for $$H'(12)$$, we should average the rates of change from the intervals on either side of $$t=12$$.

The backward difference (rate on $$[10, 12]$$) is $$ \frac{H(12)-H(10)}{12-10} = \frac{24-20}{2} = 2$$.

The forward difference (rate on $$[12, 15]$$) is $$ \frac{H(15)-H(12)}{15-12} = \frac{33-24}{3} = 3$$.

The best estimate is the average of these two rates: $$ \frac{2+3}{2} = 2.5$$ cm/day.

The rate of change of the temperature at $$t=5$$ can be estimated by the average rate of change over the interval $$[5, 5.5]$$.

$$T'(5) \approx \frac{T(5.5) - T(5)}{5.5 - 5} = \frac{48 - 40}{0.5} = \frac{8}{0.5} = 16$$ degrees Celsius per minute.

We are given $$D(10) = 50.8$$, $$D(12) = 50$$, and $$D(14) = 49.6$$. The best estimate for $$D'(12)$$ uses the symmetric interval around $$t=12$$, which is $$[10, 14]$$.

$$D'(12) \approx \frac{D(14) - D(10)}{14 - 10} = \frac{49.6 - 50.8}{4} = \frac{-1.2}{4} = -0.3$$ feet per hour.

Acceleration is the derivative of velocity, so we need to estimate $$a(2) = v'(2)$$. We can use the average rate of change of velocity over the interval $$[2, 2.1]$$.

$$a(2) \approx \frac{v(2.1) - v(2)}{2.1 - 2} = \frac{10.5 - 10}{0.1} = \frac{0.5}{0.1} = 5$$ m/s$$^2$$.

The derivative at a point can be estimated by the slope of the secant line through two nearby points. The best estimate for $$f'(3)$$ using the given values is the average rate of change over the interval $$[2.9, 3.1]$$.

$$f'(3) \approx \frac{f(3.1) - f(2.9)}{3.1 - 2.9} = \frac{6.2 - 5.8}{0.2} = \frac{0.4}{0.2} = 2$$.

The rate of change of the volume at $$t=2$$ can be estimated by the average rate of change over the interval $$[1.5, 2]$$.

$$V'(2) \approx \frac{V(2) - V(1.5)}{2 - 1.5} = \frac{300 - 280}{0.5} = \frac{20}{0.5} = 40$$ gallons per hour.

To find the best estimate for $$g'(2)$$, we should use the smallest symmetric interval around $$x=2$$. The values at $$x=1.9$$ and $$x=2.1$$ provide this interval.

$$g'(2) \approx \frac{g(2.1) - g(1.9)}{2.1 - 1.9} = \frac{4.9 - 4.2}{0.2} = \frac{0.7}{0.2} = 3.5$$.

The derivative $$f'(4)$$ can be approximated by the slope of the secant line through the points at $$x=3.99$$ and $$x=4.01$$.

$$f'(4) \approx \frac{f(4.01) - f(3.99)}{4.01 - 3.99} = \frac{6.277 - 6.431}{0.02} = \frac{-0.154}{0.02} = -7.7$$.

The best estimate for $$h'(5.0)$$ can be found by using the symmetric difference quotient over the smallest symmetric interval $$[4.8, 5.2]$$. This is equivalent to averaging the backward and forward difference quotients.

$$h'(5.0) \approx \frac{h(5.2) - h(4.8)}{5.2 - 4.8} = \frac{11.6 - 10.2}{0.4} = \frac{1.4}{0.4} = 3.5$$.

The rate of population growth at $$t=9$$ is $$P'(9)$$. We can estimate this value using the average rate of change over the interval $$[8, 10]$$, for which $$t=9$$ is the midpoint.

$$P'(9) \approx \frac{P(10) - P(8)}{10 - 8} = \frac{162 - 150}{2} = \frac{12}{2} = 6$$. This is 6 thousand people per year.