Grade is not equal to injective dimension Let $R$ be commutative Noetherian ring but not necessary local ring, and $I$ be proper ideal of $R$. I want to find an example of ring such that
$\operatorname{Ext}_R^i(R/I,R)\neq 0$ is not zero at least in two point and finite time. on the other hand $\operatorname{grade}(R/I,R)\neq \operatorname{injdim}(R)$, and $\operatorname{injdim}(R)$ is finite.
 A: https://math.stackexchange.com/questions/3459133/quotient-of-a-local-cohen-macaulay-ring-by-a-minimal-prime gives example of a local Gorenstein ring $R$ with a minimal prime ideal $P$ such that $R/P$ is not Cohen-Macaulay. Let $n= \text{depth } R - \text{depth } (R/P)$. Then $$n=\dim R - \text{depth}(R/P)> \dim R - \dim (R/P)\ge 0$$
Since $R$ is Gorenstein so $R$ has finite injective dimension, hence $\text{Ext}^n_R(R/P,R)\ne 0$ by Ischebeck's Theorem (see Theorem 11.2 of https://www.ams.org/bookstore-getitem/item=surv-181 for example). Moreover, since $P$ is an associated prime of $R$, so $R/P$ injects into $R$, so $(\text{Ext}^0_R(R/P,R)\cong ) \text{Hom}_R(R/P,R)\ne 0$, so $\text{grade}(R/P,R)=0<\dim R=\text{inj.dim } R$. As $n>0$, so we have got non-vanishing of $\text{Ext}^i_R(R/P,R)$ for at least two distinct values of $i$. As $R$ has finite injective dimension, so clearly $\text{Ext}^{>\dim R}_R(R/P,R)=0$. Thus $R$ and $P$ satisfies all your requirements.
Update: Let $(R,\mathfrak m)$ be a Gorenstein local ring of positive dimension, and let $I$ be a non-zero ideal with $\text{ht}(I)< \dim R$. Then, $R$ and $I\mathfrak m$ satisfy all your requirements.
Indeed, first let me show that $R/I\mathfrak m$ has depth $0$: By Nakayama's Lemma $I\mathfrak m \subsetneq I$ which is contained in $(I\mathfrak m:_R\mathfrak m)$ consequently, $\text{Hom}_R(R/\mathfrak m, R/I\mathfrak m)\cong \dfrac{(I\mathfrak m:_R\mathfrak m)}{I\mathfrak m } \ne 0 $, showing $R/I\mathfrak m$ has depth $0$. Hence, $\text{depth } (R)-\text{depth }(R/I\mathfrak m)=\dim R$, so $R$ Gorenstein now implies $\text{Ext}_R^{\dim R}(R/I\mathfrak m,R)\ne 0$ by Ischebeck's Theorem. As $R$ is Cohen-Macaulay, we also get $$\text{grade}(R/I\mathfrak m,R)=\text{ht}(I\mathfrak m)\le \text{ht}(I)<\dim R=\text{inj.dim } R$$ Hence $\text{Ext}^i_R(R/I\mathfrak m, R)$ vanishes for at least two distinct values of $i$ namely $\dim R$ and $\text{grade}(R/I\mathfrak m,R)$.
