It is given that
$\frac{3 x^4+5 x^2+2}{\left(x^2+1\right)^2\left(x^2+2\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+1}+\frac{E x+F}{\left(x^2+1\right)^2}$
$\begin{aligned} \Rightarrow\left(3 x^4\right. & \left.+5 x^2+2\right)=(A x+B)\left(x^2+1\right)^2 \\ & +(C x+D)\left(x^2+1\right)\left(x^2+2\right)+(E x+F)\left(x^2+2\right)\end{aligned}$
On putting $x^2=-1$, we get
$3-5+2=(i E+F)(1) \Rightarrow E=0, F=0$
On putting $x^2=-2$, we get
$12-10+2=(\sqrt{2} i A+B)(-1)^2$
$\begin{array}{ll}\Rightarrow & 4=\sqrt{2} i A+B \\ \Rightarrow & A=0, B=4\end{array}$
Now, on putting $x=0$, we get
$2=B+2 D+2 F$
$\Rightarrow \quad 2 D=-2 \Rightarrow D=-1$
So,
$A+2 B+D+4 E$
$\begin{aligned} & =0+2(4)+(-1)+4(0) \\ & =8-1=7\end{aligned}$
Hence, option (d) is correct.