You can basically think of this quotient ring as "$\mathbb{Q}$ with $\sqrt{-1}=x$ adjoined", or "the complex numbers with rational coefficients". (Fun to say.)
That said, #1 obviously has $x$ as a root.
For 2 and 3, you could check to see what their real roots look like with the quadratic formula... if those roots are in $R$, then they're reducible.
For 4 at the very worst you could actually substitute $ax+b$ and solve the resulting system of equations to see if you can get a solution in your extension. Or you can apply the rational root test and see if it has any rational roots immediately.
I get "no, yes, yes, no"