Finding $\text{Ext}_R^n(M,R)$ and $\text{Ext}_R^n(M,N)$ for a particular case

1. Like you said, $0\to R \xrightarrow{\cdot 2x_2} R \to M\to 0$ is a free resolution $M$, therefore $\mathrm{Ext}^k(M,R)$ is the $k$-th homology group of the sequence $0 \to\mathrm{Hom}(R,R) \xrightarrow{\cdot 2x_2} \mathrm{Hom}(R,R)\to 0$. Hence :

$\mathrm{Ext}^0(M,R)=\frac{\ker(\cdot 2x_2:\mathrm{Hom}(R,R)\to \mathrm{Hom}(R,R))}{\mathrm{im}(0: 0\to \mathrm{Hom}(R,R))}=\frac{\ker(\cdot 2x_2 : R\to R)}{(0)}=0$

$\mathrm{Ext}^1(M,R)=\frac{\ker(\mathrm{Hom}(R,R)\to 0)}{\mathrm{im}(\cdot 2x_2: \mathrm{Hom}(R,R)\to \mathrm{Hom}(R,R))}=\frac{\ker( 0: R\to 0)}{\mathrm{im}(\cdot 2x_2: R\to R)}=R/(2x_2)=M$

and $\mathrm{Ext}^k(M,R) =0$ for $k\ge 2$.

2. Similarly, we can use the sequence $0 \to \mathrm{Hom}(R,N) \xrightarrow{\cdot 2x_2} \mathrm{Hom}(R,N)\to 0$ to compute $\mathrm{Ext}^k(M,N)$. We find :

$\mathrm{Ext}^0(M,N)=\frac{\ker(\cdot 2x_2:\mathrm{Hom}(R,N)\to\mathrm{Hom}(R,N))}{\mathrm{im}(0: 0\to\mathrm{Hom}(R,R))}=\ker(\cdot 2x_2 : N\to N) = \mathbb{Z}[x_1]$ (seen as the $R$-module $\mathbb{Z}[x_1,x_2]/(x_2)$)

$\mathrm{Ext}^1(M,N)=\frac{\ker(\mathrm{Hom}(R,N)\to 0)}{\mathrm{im}(\cdot 2x_2: \mathrm{Hom}(R,N)\to\mathrm{Hom}(R,N))}=\frac{\ker( 0: N\to 0)}{\mathrm{im}(\cdot 2x_2: N\to N)}=N/(2x_2 N)$

and $\mathrm{Ext}^k(M,N)=0$ for $k\ge 2$.