Characteristic of the basic field is zero in this question. Let $E \subset \mathbb{P}^2$ be a smooth elliptic curve. Let $\mathcal{F}$ be the vector bundle on $E$ obtained as restriction to $E$ of the sheaf of endomorphisms of $T_{\mathbb{P}^2}$ $$ \mathcal{F}=\mathcal{End}(T_{\mathbb{P}^2})\big|_E. $$ I want to compute dimensions of cohomology groups of $\mathcal{F}$. First of all, Riemann-Roch formula for elliptic curve says $\operatorname{deg}(\mathcal{F})=\chi(\mathcal{F})$. Noting that $$ \mathcal{F} \cong T_{\mathbb{P}^2}|_E \otimes T_{\mathbb{P}^2}^\vee \big|_E $$ we get $\operatorname{deg}(\mathcal{F})=0$, therefore $\operatorname{dim}H^0(E,\mathcal{F})=\operatorname{dim}H^1(E,\mathcal{F})$.
Next observation. The trace map $\operatorname{tr}:\mathcal{End}(T_{\mathbb{P}^2}) \to \mathcal{O}_{\mathbb{P}^2}$ gives a short exact sequence $$ 0 \to \operatorname{ker}(tr) \to \mathcal{End}(T_{\mathbb{P}^2}) \to \mathcal{O}_{\mathbb{P}^2} \to 0, $$ that always splits. It means that vector bundle $\mathcal{F}$ is decomposible and have a summand $\mathcal{O}_E$. Therefore $\operatorname{dim}H^0(E,\mathcal{F})=\operatorname{dim}H^1(E,\mathcal{F}) \geq 1$.
But how actually compute this dimensions?