jmc

Since $a$ and $b$ are roots of $x^2-mx+2=0,$ we have $$x^2-mx+2=(x-a)(x-b)\text{and}ab=2.$$ In a similar manner, the constant term of $x^2-px+q$ is the product of $a+(1/b)$ and $b+(1/a),$ so $$q=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right)=ab+1+1+\frac{1}{ab}=\boxed{\frac{9}{2}}.$$