(a) Find the polynomials $f(x)$ and $g(x)$ with integer coefficients such that $$ \dfrac{f(\sqrt 3 + \sqrt 5)}{g(\sqrt 3 + \sqrt 5)} = \sqrt 3 $$ (b) Find $f$ and $g$ so that $$ \dfrac{f(\sqrt 3 + \sqrt 5)}{g(\sqrt 3 + \sqrt 5)} = \sqrt 5 $$
I can't think of a way to solve this problem.
Just take $f(x)=x^2-2$, $g(x)=2x$ for part (a) and $f(x)=x^2+2$, $g(x)=2x$ for part (b). It's not hard to check that $f, g$ are satysfying all conditions.
However, how we can find such polynomials? Firstly, you can try to find $f, g$ in form $f(x)=a_1x^2+b_1x+c_1$ and $g(x)=a_2x^2+b_2x+c_2$ by solving corresponding system of linear equations.
There is also another way. Note that the condition of part (a) is equivalent to $h(\sqrt{3}+\sqrt{5})=0$, where $h(x)=f(x)-\sqrt{3}g(x)$. Now, define new function $\varphi$ which maps $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ to $a+b\sqrt{3}-c\sqrt{5}-d\sqrt{15}$, where $a,b,c,d\in\mathbb{Z}$ (in other words, $\varphi$ change $\sqrt{5}$ into $-\sqrt{5}$). Since $f$ and $g$ has integer coefficients we have $$ f(\varphi(x))=\varphi(f(x)), g(\varphi(x))=\varphi(g(x)) $$ if $x$ has form $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$, where $a,b,c,d\in\mathbb{Z}$. Hence, $$ h(\sqrt{3}+\sqrt{5})=f(\sqrt{3}+\sqrt{5})-\sqrt{3}g(\sqrt{3}+\sqrt{5})=\varphi(f(\sqrt{3}-\sqrt{5}))-\sqrt{3}\varphi(g(\sqrt{3}-\sqrt{5})). $$ Moreover, $$ \varphi(f(\sqrt{3}-\sqrt{5}))-\sqrt{3}\varphi(g(\sqrt{3}-\sqrt{5}))=\varphi(f(\sqrt{3}-\sqrt{5})-\sqrt{3}g(\sqrt{3}-\sqrt{5}))=\varphi(h(\sqrt{3}-\sqrt{5})). $$ Thus, $\varphi(h(\sqrt{3}-\sqrt{5}))=0$, so $h(\sqrt{3}-\sqrt{5})=0$. It means that $\sqrt{3}\pm\sqrt{5}$ are roots of $h(x)$, so $h(x)$ is divisible by $(x-\sqrt{3}+\sqrt{5})(x-\sqrt{3}-\sqrt{5})=(x-\sqrt{3})^2-5=x^2-2\sqrt{3}x-2$.
At this point we can simply take $h(x)=x^2-2\sqrt{3}x-2$. Since $h(x)=f(x)-\sqrt{3}g(x)$ we obtain $f(x)=x^2-2$ and $g(x)=2x$. Analgously, we can get example for part (b).