If $T$ is a closed operator with non-empty resolvent set $\rho(T)$ and you take $\alpha\in \rho(T)$, then $A:=(T-\alpha I)^{-1}$ is in the Banach algebra of bounded operator on $X$, and given an analytic function $f$ on a neighborhood of $\sigma(T)$ you can find an analytic function $g$ on a neighborhood of $\sigma(A)$ so that $f(T)=g(A)$.

For finite sums and a closed operator $T$ acting on a complex Banach space $X$, the formula is valid in general. This is a consequence of the properties of the analytic functional calculus. See Theorem 9.1 in Chapter V of the book A.E. Taylor and D.C. Lay "Introduction to functional analysis, 2nd ed". Wiley 1980.

I was thinking about a practical criterium. For example, it obviously follows from the Gelfand-Naimark theorem that the space $Bo[0,1]$ of bounded Borel functions on $[0,1]$ is a $C(K)$ space. Moreover I am not familiar with Hopf algebras and Kac algebras.

Can $L^1(G)$ be isomorphic to $C(\hat G)$ when $G$ is infinite discrete?

To Bill Johnson: There are trivial cases in which it is not. Say a finite set in which some of the points have finite measure and some others infinite measure.

A concrete answer to Q$_3$: every surjective space is isomorphic to $\ell_1(\Gamma)$, which is a dual space. However there are injective spaces not isomorphic to $\ell_\infty(\Gamma)$; even not isomorphic to a dual space. See Corollary 4.4 in [Rosenthal; Acta Math. 124 (1970), 205-248].