jmc

Since $(x-1)(x-2)$ has degree $2$, we know the remainder has degree at most $1$, and thus is of the form $ax+b$ for some constants $a$ and $b$. Let $q(x)$ be the quotient. Then we have, $$f(x)=(x-1)(x-2)q(x)+ax+b.$$Substituting $x=1$ and $x=2$ gives us the equations: $$\begin{array}{rl}f(1)& =2=a+b\\ f(2)& =3=2a+b\end{array}$$Subtracting the first equation from the second gives us $a=1$ and consequently, $b=1$. Hence the remainder is $\boxed{x+1}$.