1) Assume you're in an open set $U\subset\mathbb{R}^n$ then the mapping $u\mapsto (u,\nabla u)$ from either $H^1$ or $H_0^1$ to $L^2\times(L^2)^n$ equipped with the appropiate product norm gives an isometry. Since $H^1$ and $H_0^1$ are Banach spaces, they're also closed subspaces of a reflexive space and so are reflexive themselves.
2) By the Poincaré inequality (the one that says $\| u\|_{H_0^1}\leq c\| \nabla u \|_{L^2}$ for every $u\in C_c^\infty$) the inner product $(\nabla u , \nabla v)_{L^2}$ is equivalent to the usual inner product in $H_0^1$ so the answer is yes.