jmc

Evaluating $f(x)$ for $x=3$ and $x=-3$, we have $$\{\begin{array}{rl}f(3)& =a\cdot 3^4-b\cdot 3^2+3+5,\\ f(-3)& =a\cdot (-3{)}^4-b\cdot (-3{)}^2+(-3)+5.\end{array}$$If we subtract the second equation from the first equation, all the terms but one cancel out, and we get $$f(3)-f(-3)=3-(-3)=6.$$Thus, if $f(-3)=2,$ then $f(3)=f(-3)+6=2+6=\boxed{8}.$