1. The polynomial $p_{1}=x-a$ can have a root $x=a$ coinciding with one of the roots of the product $p(x)=p_{1}(x) \cdot p_{2}(x)$.

Case $1 \quad a=1$

Then the polynomial $p_{2}(x)=(x-1)^{r}(x-2)^{s}(x+3)^{t}$, where $r \geq 1, s \geq 1, t \geq 1-$ are integers, $r+s+t=4$, satisfies the condition of the problem. The polynomial sum $p_{1}(x)+p_{2}(x)$ has a constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}-1=(-1)^{t} \cdot 2^{s} \cdot 3^{t}-1$ which has a maximum value $u_{\max }=17$ when $r=1, s=1, t=2$.

Case $2 \quad a=2$

Then the polynomial $p_{2}(x)=(x-1)^{r}(x-2)^{s}(x+3)^{t}$ and the polynomial $p_{1}(x)+p_{2}(x)$ has a constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}-2=(-1)^{t} \cdot 2^{s} \cdot 3^{t}-2$, taking the maximum value $u_{\max }=16$ when $r=1, s=1, t=2$.

Case $3 \quad a=-3$

The constant term $u=(-1)^{r+s} \cdot 2^{s} \cdot 3^{t}+3$ takes the maximum value $u_{\max }=21$ when $s=r=1, t=2$

The required polynomials: $p_{2}(x)=(x-1)(x-2)(x+3)^{2}, p_{1}(x)=x+3$
Answer: $p_{1}(x)=x+3, p_{2}(x)=(x-1)(x-2)(x+3)^{2} ; a_{0}=21$