I am wanting to solve part d. For c, I have the line's equation as $y=-3x+\cfrac{57}{5}$. I am not sure if this is the correct answer.
How can I find the zero indicated in part d if I do not know how to find the equation of the tangent line for x=4? Do I just use x=3?
For part c, use the equation of the tangent line at $(3,F(3))$: $$y-F(3)=F'(3)(x-3)$$
$$ F'(x)=\left(\int_{1}^{x}f(t)\,dt\right)'=f(x)\implies F'(3)=f(3)=-3 $$
The fact that $f(3)=-3$ follows from the graph.
The last piece of information that we need to find is the value of $F(3)$:
$$F(3)=\int_{1}^{3}f(t)\,dt$$
Again, we're going to use the graph. Notice that on the interval $[1,2]$ $f(t)=2$ and on the interval $[2,3]$ $f(t)=-5t+12$ (given the two points $(2,2)$ and $(3,-3)$, it's not difficult to find the equation of that line):
$$F(3)=\int_{1}^{3}f(t)\,dt=\int_{1}^{2}2\,dt+\int_{2}^{3}(-5t+12)\,dt=\\ 2t\bigg|_{1}^{2}-\left[\frac{5t^2}{2}-12t\right]_{2}^{3}= 2\cdot(2-1)-\left[\frac{5\cdot 3^2}{2}-12\cdot 3-\left(\frac{5\cdot 2^2}{2}-12\cdot 2\right)\right]=2-\frac{1}{2}=\frac{3}{2} $$
So, the equation of the tangent line to the curve $y=F(x)$ at $(3,F(3))$ is going to be: $$y-\frac{3}{2}=-3(x-3)\implies y=-3x+\frac{21}{2}$$
It's generally assumed that locally a tangent line to a curve at a particular point very well approximates that curve at that point. This idea comes from the fact that near the point of tangency, the tangent line and the curve geometrically lie very close to each other and for approximation purposes we can assume that they're actually equal. So, near the point $x=3$, we can use the line equation that we came up with to find an approximate value for $F(x)=0$:
$$0=-3\cdot x+\frac{21}{2}\implies x=\frac{7}{2}$$
This means that if you take $x=\frac{7}{2}$ and plug it into $F(x)$, the value that you'll get is going to be very close to zero.