9. Write the quadratic trinomial $x^{2}+9 x+47$ on the blackboard. Tanya arbitrarily increases or decreases the coefficient of $x$ by 1, and then Vanya increases or decreases the constant term by a fixed number $m$, and then the operations are repeated. If the polynomial written on the blackboard has an integer root, Vanya gets 5 points. If (1) $m=2,(2) m=3$, please discuss separately whether Vanya can be guaranteed to get 5 points under any operations.
9. (1) Yes; (2) No.
(1) Vanya's operations are independent of Tanya's, and in a finite number of steps (either his own or Tanya's), he can ensure that the value of the quadratic trinomial at the point \( x=1 \) is 0.

Initially, this value is \( 1+9+47=57 \). Next, Tanya can change this value by 1 each step, and Vanya can change it by 2 each step. Therefore, Vanya can ensure that after every two steps, this value gets closer to 0, at least until Tanya's next step, when its absolute value is no more than 2. If Tanya makes it 0, Vanya achieves his goal; if it is 1 or -1, Vanya can make it -1 or 1, respectively, and after Tanya's next step, it will become 0, 2, or -2. In any case, Vanya can achieve his goal.
(2) Tanya's strategy is to make the coefficient of the \( x \) term equal to 7 or 8. In this case, no matter how Vanya operates, the resulting quadratic trinomial will not have integer roots.

In fact, after each of Tanya's steps, the trinomial for any integer \( n \) will have the form
\[
x^{2}+7 x+47-3 n \text { or } x^{2}+8 x+47-3 n \text {. }
\]

Since these values are not divisible by 3 (and particularly not equal to 0), these two polynomials do not have integer roots.