jmc

By the Remainder Theorem, $p(1)=3,$ and $p(3)=5.$

Let $q(x)$ and $ax+b$ be the quotient and remainder, respectively, when the polynomial $p(x)$ is divided by $(x-1)(x-3),$ so $$p(x)=(x-1)(x-3)q(x)+ax+b.$$Setting $x=1,$ we get $p(1)=a+b,$ so $a+b=3.$ Setting $x=3,$ we get $p(3)=3a+b,$ so $3a+b=5.$ Solving, we find $a=1$ and $b=2.$

Therefore, the remainder is $\boxed{x+2}.$