For parts a and b, go back and look up the Intermediate Value Theorem, if you have a calc textbook. Or google it. Basically you need to find h(1) and h(3), then show that -5 is in between these two values.
For part b, you need one additional step. You will have to calculate h'(1) and h'(3) and then show that -5 is in between those values. But to calculate h'(c), use the h function you were given: h(x) = f(g(x)) - 6. Don't forget the chain rule, since f and g are both functions. Then plug in values from your table.
Part c: You need to go back to a more basic understanding of f ''. f '' would be the derivative of f ', or the slope of f '. "Estimate" usually means that you don't have to use calculus, or that they want you to algebraically calculate a derivative by finding the slope over an interval. So to find the "slope" of f ' when x is 2.5, use the algebraic equation for slope: (y1 - y2)/(x1 - x2). Your two points would be the points closest to 2.5, or from your chart use x = 2 and x = 3 and the corresponding values of f ' , which are 2 and -4. To summarize: find the slope of f ' over the interval from 2 to 3 as an estimate for the slope at 2.5. If you are using a calc textbook, this would be similar to the beginning of your course before you knew how to calculate a derivative using calculus. Somewhere close to "definition of derivative."
Part d: You need to know that, essentially, inverse functions have the x and y values switched. So when you are asked for the line tangent to y = g^-1(x) at x = 2, you are really looking for the point on g where the y is 2. From your table, g(x) is 2 when x is 1. So that means for g^-1(x), when x is 2, y is 1. That's your point: (2,1). Now a tangent line would have the same slope as its curve at that point. So you need the derivative of g^-1(x) at the point (2,1). Inverse functions have reciprocal slopes, so the derivative of g^-1(x) at (2,1) is the reciprocal of the slope of g(x) at that point. From your table, g'(x) is 5, so your slope is 1/5. Now you can use m = 1/5 and the point (2,1) and the point slope equation of a line to write your tangent line equation.
I know that doesn't spell out the answer for you, but hopefully it gets you started and on your way to the answer. Wouldn't want to do your homework. ;-)
I teach AP Calculus.