well, about A and D, I just plugged the values on the slope formula of

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. 
for D the values are -2.25 and 8.0, so no dice on that one.
for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.
checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. 
for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.
for E, do they intercept at (2,1)? well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. 
for F, we know the slope of the line L(x) is 8, is g'(2) = 8? let's check
recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. 