Well-known Theorem:

Let $a$ be an ideal of the noetherian ring $R$ and let $M$ be a finitely generated $R$-module. Let $i \in \Bbb N_0$ be such that $H^j_a (M)$ is finitely generated for all $j < i$. Then the set $Ass_R((H^i_a (M))$ is finite.

Now my questions:

Is there an example that $H^0_a(M), H^1_a(M), H^2_a(M)$ be **non-zero** finitely generated modules.
Is there an example that $H^0_a(M), H^1_a(M),..., H^j_a(M)$ be **non-zero** finitely generated modules for $j>2$.

**Example for** $\bf{H^0_a(M), H^1_a(M)}$:

Let $S=k[x_1,x_2,x_3,x_4]$ be a polynomial ring over field $k$ and $I=P_1\cap P_2$ where $P_1=(x_1,x_2), P_2=(x_3,x_4)$ and $m=(x_1,x_2,x_3,x_4)$. Set $R=S/I$. From the exact sequence $$0\rightarrow R\rightarrow S/P_1\oplus S/P_2\rightarrow S/m\rightarrow 0$$ we have the exact sequence $$0\rightarrow H^0_m(R)\rightarrow H^0_m(S/P_1)\oplus H^0_m(S/P_2)\rightarrow H^0_m(S/m)\rightarrow H^1_m(R)\rightarrow H^1_m(S/P_1)\oplus H^1_m(S/P_2)\rightarrow H^1_m(S/m)\rightarrow H^2_m(R)\rightarrow H^2_m(S/P_1)\oplus H^2_m(S/P_2)\rightarrow H^2_m(S/m)$$ Since $S/m$ is $m$-torsion then $H^i_m(S/m)=0$ for all $i\neq 0$. on the other hand $H^i_m(P_j)=0$ for $i=0,1 ,j=1,2$. Hence $H^0_m(R)=0,~ H^1_m(R)\cong H^0_m(S/m)=S/m$ are finitely generated.