Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative).
Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the left and from the right (I'd call this an $\left(R,R\right)_k$-bimodule, but I haven't seen this notation anywhere). There are two ways to define the Hochschild homology and cohomology of $R$ with coefficients in $M$: either as the homology of the standard complex tensored with $M$ rsp. the cohomology of Hom of the standard complex and $M$, or as $\mathrm{Tor}$ and $\mathrm{Ext}$. As far as I understand, these two definitions are equivalent only if $R$ is a projective $k$-module, which I don't want to require here.
Question 1: So let me define Hochschild cohomology and homology through the standard complex. Then, Löfwall's text seems to silently hint at the fact that if a $k$-algebra $R$ satisfies $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then it also satisfies $\mathrm{H}_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$. While this is clear from homological algebra in the case when $R$ is a projective $k$-module, is this true otherwise? And how is it proven?
(Remark: A $k$-algebra $R$ such that $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$ is said to be zero-dimensional (in Löfwall's text) or separable (in the modern sense of this word).)
Question 2: The same text gives a counterexample for the opposite direction (if $\mathrm{H}_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$). In this counterexample, $k$ is a field and $R$ is commutative, but infinite-dimensional. I assume that counterexamples fade when we impose some more restrictive conditions on $k$ and $R$. What about finite-dimensional $R$? What about finitely-generated-as-algebras $R$? If $k$ is not a field anymore? If $R$ is not commutative anymore?