Since the point of tangency is on both f(x) and the tangent line,
and since when x=4, the tangent line and x=4 has y value 

y = -5(4)+6 = -20+6 = -14, 

then (4,-14) the point on tangency is a point on the 
graph of f(x).  That is,

f(4) = -14 

Since f'(4) is the slope of the tangent line at x=4, and the tangent
line is y = -5x+6 which has slope -5 (which is found either by 
comparing it to y = mx+b, or by finding ). 
then f'(4) = 5

g(x) = 2f(x)+8

so

g(4) = 2f(4)+8
g(4) = 2(-14)+8
g(4) = -28+8
g(4) = -20

Since 

g(x) = 2f(x)+8,

taking the derivative of both sides:

g'(x) = 2f'(x)
g'(4) = 2f'(4)
g'(4) = 2(5)
g'(4) = 10

Edwin