Projective variety with $H^i(X, \bigwedge^j T_X) \cong 0$ and $H^i(X, \Omega^j_X) \cong 0$ Let $X$ be a smooth projective variety over a field. I know that
$$
H^i(X, \bigwedge^j T_X) \cong 0,
$$
for $i+j >0$ and also
$$
H^i(X, \Omega^j_X) \cong 0,
$$ 
for $i \neq j$.
Is it enough to conclude that $X$ is a point?
 A: This is not an answer, but I am curious as to how you came to this question. It looks interesting and I would love to see a complete answer. Here, I just wanted to give an argument for curves (easy) and surfaces (not so easy).
For curves, $H^0(\Omega^1_X=K_X)=0$ implies it is rational and then $H^0(T)=H^0(-K)\neq 0$. So, curves satisfying your conditions do not exist.
For surfaces, you have $q=h^1(\mathcal{O})=0, p_g=h^2(\mathcal{O})=0$. You also have $0=h^2(\wedge^2T_X=-K_X)$ which by Serre dulaity implies $h^0(2K_X)=0$. These together imply by Castelnuovo's theorem, that the surface is rational. Since $h^i(\wedge^2T_X)=h^i(-K)=0$ for all $i$, by Riemann-Roch, we have $0=\chi(-K)=\frac{-K(-2K)}{2}+1$ and thus $K^2=-1$. By Noether's formula, we also have $c_1^2+c_2=12$ and since $c_1^2=-1$, $c_2=13$. Finally, we use the fact $h^i(T_X)=0$ for all $i$ and thus, by Riemann-Roch, $0=\chi(T_X)=\frac{-K(-2K)}{2}-c_2+2=-c_2+1$, giving us $c_2=1$, contradicting the earlier equation $c_2=13$. So, such surfaces can not exist.
May be there is a clean argument for all dimensions, and hope some one posts a complete answer.
