$\operatorname{Tr}(AB)=\operatorname{Tr}(A)\operatorname{Tr}(B)$ 
Problem. Show that $\operatorname{Tr}(AB)=\operatorname{Tr}(A)\operatorname{Tr}(B)$, if $A^2+3AB+B^2=BA$ and $\det(A)=0$, where all matrices involved is real valued $2\times2$ matrices.

First I rewrote the $A^2+3AB+B^2=BA$ as 
$$ A(A+3B)=B(A-B) $$
and I took the determinant. Then I obtained that
$$ \det(B)\det(A-B)=0, $$
because $\det(A)=0$. So, $\det(B)=0$ or $\det(A-B)=0$.
Case I. Then, by Cayley-Hamiton theorem, I wrote that
$$ A^2=(\operatorname{Tr}(A))A \text{ and } B^2=(\operatorname{Tr}(B))B. $$
Multiplying the above two equations, I got $(AB)^2=(\operatorname{Tr}(A)\operatorname{Tr}(B))AB$. Another use of Cayley theorem got $(AB)^2=(\operatorname{Tr}(AB))AB$. So, $\operatorname{Tr}(A)\operatorname{Tr}(B)$ must be equal with $\operatorname{Tr}(AB)$ only if $AB=BA$. But I don't have such a hypothesis.
Case II. For the case where $\det(A-B)$ I don't know how to handle this one in order to get the answear.

Sangchul Lee's answer can be slightly shortened as follows.
First of all, the key observation is that
$$
(A+B)^2 = 2[B,A] = 2[A+B,A],\tag{1}
$$
where the square bracket denotes a commutator.
By Jacobson's lemma, if $X$ is an $n\times n$ matrix over some field whose characteristic is either zero or $>n$, and $X$ commutes with some commutator $[X,Y]$, the commutator $[X,Y]$ must be nilpotent. So, in our case, if the characteristic of the underlying field is not $2$, since $A+B$ commutes with $(A+B)^2=2[A+B,A]$, Jacobson's lemma implies that $[A+B,A]$ is nilpotent. Hence $(A+B)^2$ and in turn $A+B$ are nilpotent and $(A+B)^2=0$. But then $(1)$ also implies that $[B,A]=0$, i.e. $A$ and $B$ commute.
Let $E$ be the algebraic closure of the field generated by the eight matrix entries in $A$ and $B$. Since $A$ and $B$ commute, they are simultaneously triangulable over $E$. This, together with the facts that $A$ is singular and $A+B$ is nilpotent, imply that $B$ must have a zero eigenvalue, i.e. $B$ is singular.
Let $a=\operatorname{tr}(A)$ and $b=\operatorname{tr}(B)$. On one hand, as both $A$ and $B$ are singular, by Cayley-Hamilton theorem we have
\begin{align}
\operatorname{tr}((A+B)^2)
&=\operatorname{tr}(A^2+2AB+B^2)\\
&=\operatorname{tr}(aA+2AB+bB)\\
&=a^2+2\operatorname{tr}(AB)+b^2.\tag{2}
\end{align}
On the other hand,
\begin{align}
(\operatorname{tr}(A+B))^2
&=(\operatorname{tr}(A)+\operatorname{tr}(B))^2\\
&=a^2+2\operatorname{tr}(A)\operatorname{tr}(B)+b^2.\tag{3}
\end{align}
As $A+B$ is nilpotent, both $\operatorname{tr}((A+B)^2)$ and $(\operatorname{tr}(A+B))^2$ are zero. Now the result follows by equating $(2)$ and $(3)$.
Finally, note that the assumption that characteristic of the field is not $2$ is essential here. E.g. consider
$$
A=\pmatrix{0&0\\ 1&0},\ B=\pmatrix{1&1\\ 0&1}.
$$
If the ground field is $GF(2)$, the conditions that $A^2+3AB+B^2=BA$, or equivalently $(A+B)^2=0$ (because $1+1=0$), and that $\det(A)=0$ are both satisfied, but clearly $\operatorname{tr}(AB)=1\ne0=\operatorname{tr}(A)\operatorname{tr}(B)$ here.
