a) Let $p(x)\in \mathbb{R}[x]$. Define $\Phi(p)=(p(2),p(3))$. Then for a constant polynomial $p=a$ we have $\Phi(a)=(a,a)$ and for $p=x$ we have $\Phi(x)=(2,3)$. We also have $$\Phi(p+q)=(p(2)+q(2),p(3)+q(3))=(p(2),p(3))+(q(2),q(3))=\Phi(p)+\Phi(q)$$ and $$\Phi(pq)=(p(2)q(2),p(3)q(3))=(p(2),p(3))\cdot (q(2),q(3))=\Phi(p)\Phi(q)$$ so $\Phi$ is a ring homomorphism.

If $\Psi$ is another ring homomorphism with these properties for any polynomial $p(x)=a_0 +a_1 x+ \dots +a_n x^n$ we have $$ \Psi(p(x))=\Psi(a_0 +a_1 x+ \dots +a_n x^n)= \Psi(a_0) +\Psi(a_1)\Psi( x)+ \dots +\Psi(a_n)\Psi( x^n)\\ =\Psi(a_0) +\Psi(a_1)\Psi( x)+ \dots +\Psi(a_n)\Psi( x)^n=(a_0,a_0)+(a_1,a_1)(2,3)+\dots +(a_n,a_n)(2,3)^n \\ =(a_0,a_0)+(a_1,a_1)(2,3)+\dots +(a_n,a_n)(2^n,3^n)=(a_0,a_0)+(a_12,a_13)+\dots +(a_n2^n,a_n3^n)\\ =(a_0 +a_1 2+ \dots +a_n 2^n,a_0 +a_1 3+ \dots +a_n 3^n)=(p(2),p(3))=\Phi(p(x)) $$So $\Psi=\Phi$ and $\Phi$ is unique.

b) Let $(a,b)\in \mathbb{R}\times \mathbb{R}$. Then for $p(x)=b(x-2)-a(x-3)$ we have $p(2)=a$ and $p(3)=b$. So $\Phi(p)=(a,b)$ and $\Phi$ is surjective.

c) Suppose $p(x)$ is in the kernel of $\Phi$. Then $\Phi(p)=(0,0)$. So $p(2)=0=p(3)$. Therefore $p(x)$ has a factor of $(x-2)$ and a factor of $(x-3)$. So $p(x)=(x-2)(x-3)q(x)=(x^2-5x+6)q(x)$ for a polynomial $q(x)\in \mathbb{R}[x]$. Conversely if $p$ is of this form then it vanishes at 2,3 and therefore it is in the kernel of $\Phi$. So $$ \mathrm{Ker}\,\Phi=(x^2-5x+6)\mathbb{R}[x]$$

d) By the first isomorphism theorem $\mathbb{R}[x]/\mathrm{Ker}\,\Phi$ is isomorphic to the image of $\Phi$ which is $\mathbb{R}\times \mathbb{R}$ because $\Phi$ is surjective. So $\mathbb{R}[x]/(x^2-5x+6)\mathbb{R}[x]$ is isomorphic to $\mathbb{R}\times \mathbb{R}$.