A side note: you inverted the vector field $X$ and the 1-form $df$ in the definition of Hamiltonian vector field. $X$ is Hamiltonian if it is a Poisson gradient, and some equivalent notations are $X_{f} = \{f, \cdot\} = \pi(df, \cdot) = (df)^{\sharp} = \iota_{df}\pi $.
On a similar line, since you cannot feed vector fields to a bivector as you do with a 2-form, you cannot have a strict analogue of Cartan's magic formula for the Lie derivative of a bivector.
Back to your question
Page 122 and around there of
Paulette Libermann and Charles-Michel Marle. Symplectic Geometry and
Analytical Mechanics
can be of help without invoking Poisson cohomology, that requires building Poisson calculus first (that is, defining brackets of 1-forms from usual Poisson brackets of smooth functions).
As @studiosus said, your Poisson manifold is special in two independent ways:
A. in $\mathbb{R}^n$ differential forms are closed iff exact (all de Rham cohomology groups are trivial); and
B. the Poisson bivector is non-degenerate everywhere (making the manifold symplectic).
These implications show how these two facts make any Poisson vector field a Hamiltonian vector field:
The arrows marked with (A) or (B) hold only if the respective condition holds; let's start climbing up this chain from the bottom.
A vector field $X$ is called Poisson if $\mathcal{L}_X \pi = 0$.
The first difficulty we face is that a degenerate bivector $\pi$ defines a non-bijective homomorphism $\sharp$ mapping a 1-form into a vector field, so for a generic vector field there may not exist a 1-form $\alpha$ such that $\alpha^{\sharp} = X$.
If a Poisson vector field belongs to the image of $\sharp$, so $\mathcal{L}_X \pi = 0$ and $X = \alpha^{\sharp}$ for some 1-form $\alpha$, we say that $X$ is locally Hamiltonian.
Proposition 10.5 at page 122 of Libermann's book shows that a vector field is locally Hamiltonian iff the related 1-form is c-closed, namely $d\alpha$ vanishes on hamiltonian vector fields.
Now, if B. holds, the Poisson bivector is non-degenerate, and it defines an isomorphism between vector fields and 1-forms: in particular every vector field is associated to precisely one 1-form [* footnote].
This means that, if B. holds, every Poisson vector field is locally hamiltonian, and it can be shown that every c-closed 1-form is closed (every vector can be written as the image of a Hamiltonian vector field at one point).
To summarize, if the Poisson bivector is non-degenerate, there is a 1-to-1 correspondence between Poisson vector fields and closed 1-forms.
Finally, as said in A., in $\mathbb{R}^n$ a closed 1-form is exact: $\alpha = df$ for some function $f$; so the Poisson vector field we started with, $X = \alpha^{\sharp}$, is indeed Hamiltonian: $X = (df)^{\sharp}$.
[* footnote] If B. holds the Poisson manifold is symplectic. The symplectic form $\omega$ can be defined from the bivector since $\sharp^{-1}$ now makes sense; the non-degeneracy of $\pi$ grants the non-degeneracy of $\omega$; and interestingly the Jacobi property of the Poisson structure, or equivalently the vanishing Schouten brackets $[\pi,\pi]_S = 0$, assures that $d \omega = 0$.
Best Answer
The functions on $T^{*}Q$ that are of the form $f_{X}$ for some $X\in\mathfrak{X}(Q)$ are exactly the fiberwise linear functions $C^{\infty}_{lin}(T^{*}Q)$ on $T^{*}Q$. So the $p_{i}$ are of this form, but the $q^{j}$ are not (as they are fiberwise constant).
a) We have that $p_{i}=f_{\partial_{q^{i}}}$, since $$ p_{i}(dq^{j})=\delta_{i,j}=f_{\partial_{q^{i}}}(dq^{j}). $$ Hence, $$\label{1}\tag{1} \{p_{i},p_{j}\}=\{f_{\partial_{q^{i}}},f_{\partial_{q^{j}}}\}=-f_{[\partial_{q^{i}},\partial_{q^{j}}]}=-f_{0}=0. $$
b) As said, $q^{j}$ is not fiberwise linear, but $q^{j}p_{i}$ is. Indeed, we have $q^{j}p_{i}=f_{q^{j}\partial_{q^{i}}}$. So $$\tag{2}\label{2} \{q^{j}p_{i},p_{k}\}=\{f_{q^{j}\partial_{q^{i}}},f_{\partial_{q^{k}}}\}=-f_{[q^{j}\partial_{q^{i}},\partial_{q^{k}}]}=f_{\frac{\partial q^{j}}{\partial q^{k}}\partial_{q^{i}}}=f_{\delta_{j,k}\partial_{q^{i}}}=\delta_{j,k}f_{\partial_{q^{i}}}=\delta_{j,k}p_{i}. $$ On the other hand, using the Leibniz rule of the Poisson bracket $\{\cdot,\cdot\}$, we have $$\tag{3}\label{3} \{q^{j}p_{i},p_{k}\}=q^{j}\{p_{i},p_{j}\}+p_{i}\{q^{j},p_{k}\}=p_{i}\{q^{j},p_{k}\}, $$ using \eqref{1} in the last equality. Comparing \eqref{2} and \eqref{3} then gives $$\tag{4}\label{4} \{q^{j},p_{k}\}=\delta_{j,k}. $$
c) At last, for the fiberwise linear vector fields $q^{k}p_{i}=f_{q^{k}\partial_{q^{i}}}$ and $q^{l}p_{j}=f_{q^{l}\partial_{q^{j}}}$, we get \begin{align} \{q^{k}p_{i},q^{l}p_{j}\}&=\{f_{q^{k}\partial_{q^{i}}},f_{q^{l}\partial_{q^{j}}}\}=-f_{[q^{k}\partial_{q^{i}},q^{l}\partial_{q^{j}}]}=-f_{q^{k}\frac{\partial q^{l}}{\partial q^{i}}\partial_{q^{j}}-q^{l}\frac{\partial q^{k}}{\partial q^{j}}\partial_{q^{i}}} =-\delta_{l,i} q^{k} f_{\partial_{q^{j}}}+\delta_{k,j}q^{l}f_{\partial_{{q}^{i}}}\\ &=-\delta_{l,i} q^{k} p_{j}+\delta_{k,j}q^{l}p_{i}.\tag{5}\label{5} \end{align} But also, by the Leibniz rule: \begin{align} \{q^{k}p_{i},q^{l}p_{j}\}&=\{q^{k},q^{l}\}p_{i}p_{j}+\{q^{k},p_{j}\}p_{i}q^{l}+\{p_{i},q^{l}\}q^{k}p_{j}+\{p_{i},p_{j}\}q^{k}q^{l}\\ &=\{q^{k},q^{l}\}p_{i}p_{j}+\delta_{k,j}p_{i}q^{l}-\delta_{i,l}q^{k}p_{j},\tag{6}\label{6} \end{align} using \eqref{1} and \eqref{4} in the last equality. Comparing \eqref{5} and \eqref{6} then gives $$\tag{7}\label{7} \{q^{k},q^{l}\}=0. $$ The equalities \eqref{1},\eqref{4} and \eqref{7} now show that $\{\cdot,\cdot\}$ is indeed the canonical Poisson bracket.