Hopefully, the math gurus will help me, but I THINK that, since you're looking for the line that is perpendicular to $\displaystyle y = 1 - (\frac{x}{24})$, the slope of the line you're looking for will be the negative reciprocal of the slope of the line you have. So, for the equation of the line $\displaystyle y = 1 - (\frac{x}{24})$, you have that the slope is:
$\displaystyle -(\frac{x}{24}) = (-\frac{1}{24})x \Longrightarrow -\frac{1}{24}$
Now, the negative of this slope is $\displaystyle -(-\frac{1}{24}) \Longrightarrow \frac{1}{24}$
The reciprocal, then, is 24.
Again, let those who know correct me... ;-)
For (1)(a) the problem asked where the tangent line was perpendicular to y= 1- x/24. You found x where the derivative was 0. I have no idea why you did that! The derivative is the slope of the tangent line. A line with slope 0 is horizontal and is perpendicular to a vertical line. y= 1- x/24 is NOT vertical. It has slope -1/24. Clearly you are expected to know that if y= mx+ b and y= nx+ d are perpendicular, then n= -1/m. A line perpendicular to a line with slope -1/24 must have slope 24. THAT was why everyone set the derivative equal to 24, not 0!
For (1)(b), the problem asked where the tangent line was parallel to $\displaystyle y= \sqrt{3}- 12x$ which has slope -12. Two lines are parallel if and only if they have the same slope.
Again- two lines, y= mx+ b and y= nx+ c, are parallel if and only if m= n, perpendicular if and only if m= -1/n (equivalently mn= -1).