Lets assume f is the density.

a) The 0.8 quantile is y such that p(x<y)=0.8, or:

int(-4, y) f(x) dx=0.8.

Now we know that int(-4,0) f(x) dx=0.5, so we want y such that:

int(0,y) f(x) dx=int(0,y) {-1/16x + 14} dx =0.3

b) same as befor except that 0.8 is replaced by 0.45, also as 0.45<0.5

we are on the other bit of the definition so:

The 45th percentile is the 0.45 quantile so we seek a y such that p(x<y)=0.45, or:

int(-4, y) f(x) dx=0.45.

or

int(-4,y) f(x) dx=int(-4,y) {1/16x + 14} dx =0.45

RonL