Rom

\((f\circ g)(x) = f(g(x))\\ -\\ \begin{matrix} x &g(x) &f(g(x)) \\ 0 &4 &-6\\ 1 &3 &13 \\ \vdots \end{matrix}\\ \text{see how it works?}\)

\(f(x) = a x^3 + b x^2 + c x + d\\ f(0)= 0 \Rightarrow d = 0\\ \text{using the rest of the values we obtain 3 equations}\\ a + b + c =-5\\ -a + b -c =15\\ 8a+4b + 2c = 12\)

\(\text{adding the first two}\\ 2b = 10,~b = 5\)

\(a+c=-10\\ 8a+2c = -8\\ \text{subtract twice eq 1 from eq 2}\\ 6a = 12,~a=2\\ c= -12\)

\(f(x) = 2x^3 + 5x^2 -12x\)

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\(\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)} = \\ \dfrac{4x^2+6x}{2x^2+13x+15}\)

\(A x = y\\ Ay = x + 2y\)

\(A^2x =AAx = Ay = x+2y\)

\(A^3x=AA^2x = A(x+2y) = \\ Ax + 2Ay = y + 2(x+2y) = \\ 2x+5y\)

\(A^4x = A(2x+5y) = 2y + 5(x+2y) = 5x+12y\)

\(A^5 x = A(5x+12y) = 5y + 12(x+2y) = 12x + 29y\)

\((a,b) = (12, 29)\)

Looks like 

OOEO and OEOO

both give 7 odds in the triangle

\(\text{subtract eq 3 from eq 4}\\ y = 11\\ \text{subtract eq 2 from eq 3}\\ y + 4z + w = 4\\ 11 + 4z + w = 4\\ 4z + w = -7\)

\(\text{subtract eq 1 from eq 2}\\ y + 2z + 5w = 12\\ 11 + 2z + 5w = 12\\ 2z + 5w = 1\)

\(\text{now we have 2 equations in 2 variables}\\ w + 4z = -7\\ 5w+2z = 1\\ \text{subtract twice eq 2 from eq 1}\\ -9w = -9\\ w = 1\\ \text{and from eq 1}\\ 1 + 4z = -7\\ z = -2\)

\(\text{now dump everything we know into eq 1}\\ x + 11 + -2 + 1 = 10\\ x + 10 = 10\\ x = 0\)

\((w, x, y, z) = (1, 0, 11, -2)\)

\(\text{So someone was pulling your leg}\)

\(\text{Have you got a typo in line 4? You've got D on both sides of the scale}\)

\(\text{in (a), for example, you are told that }p(-5)=0\\ \text{thus }-5 \text{ is a zero and }x+5 \text{ is a factor}\\ \text{so divide }p(x) \text{ by }(x+5) \text{ using polynomial division}\\ \dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{this should be easy enough to factor by eye}\\ 6x^2 + x -1 = (3x -1)(2x+1) \\ \text{and it's easy enough to read off the zeros in this form as }\\ x = \dfrac 1 3,~x = -\dfrac 1 2\\ \text{follow this procedure for (b), which I leave to you}\)

btw there's a small error in your answer.. I'll let you find it!