(a)

Let $p_1(x)=x^3-2020 x^2+b_1 x+c_1$

$=(x-\alpha)(x-\beta)(x-\gamma)$

and $\quad p_2(x)=x^3-2021 x^2+b_2 x+c_2$

$=(x-\alpha)(x-\beta)(x-\delta)$

Since, $p_1(x) \cdot q_1(x)+p_2(x) \cdot q_2(x)$

$=x^2-3 x+2$

On comparing the coefficient of $x^3$, we get $q_1(x)=-q_2(x)=q(x)($ say $)$

So, $(x-\alpha)(x-\beta)[q(x)(\delta-\gamma)]$ $=(x-1)(x-2)$

$\therefore \quad \alpha=1, \beta=2, \gamma=2017$ and $\delta=2018$

$p_1(x)=(x-1)(x-2)(x-2017)$

$\Rightarrow \quad p_1(3)=-4028$

$p_2(x)=(x-1)(x-2)(x-2018)$

$p_2(1)=0$

So,$\quad p_1(3)+p_2(1)+4028=0$