$A$ is contractible if $H^1(A,X)=0$ for all Banach $A$-bimodules $X$ (here $H^1$ denotes continuous Hochschild cohomology for Banach algebras, as defined in the works of Johnson or Helemskii). It is an implicit conjecture, going back to the 1970s, that there are no "interesting" examples of contractible Banach algebras.
The confusion may have arisen in your reading, because Kamowitz's paper (which predates Johnson's 1972 monograph) only considers left $A$-modules, which he then proceeds to regard as symmetric $A$-bimodules by definiong $x\cdot a$ to be $a\cdot x$. This works because he is only considering commutative Banach algebras $A$.
So the theorem/corollary you are referring to merely says, in more modern terminology, that $H^1(C(\Omega),X)=0$ and $H^2(C(\Omega),X)=0$ whenever $X$ is a symmetric Banach $A$-bimodule. This is much weaker than requiring vanishing degree-1 cohomology for arbitary coefficient modules; if one allows arbitrary coefficient modules then there are "dimension-shift" arguments of the form $H^{n+k}(A,M) = H^n(A, V_k(A,M))$ for some new module $V_k(A,M)$ built out of $A$ and $M$ with a new, "nasty" action. (If you try to perform this construction with $A$ commutative and $M$ symmetric then $V_k(A,M)$ is usually not symmetric, and so there is no "dimension-shift" argument for cohomology with symmetric coefficients.)
Indeed, if $A$ is commutative, then the condition $H^1(A,X)=0$ for all symmetric Banach $A$-bimodules $X$ is one of the equivalent definitions of weak amenability for commutative Banach algebras, in the sense of Bade–Curtis–Dales. There are plenty of infinite-dimensional, comutative Banach algebras which are weakly amenable; anything where the linear span of idempotents is dense will do.
