**(1)** For anyons to be created locally in a physical model they must be created in groups such that the local excitation is a boson or a fermion. However, the local excitation can fractionalize into anyonic parts which can propagate independently. In terms of second quantized operators the expectation is that the the local fermionic/bosonic degree of freedom can be written as a product of anyon creation/annihilation operators. This can be explicitly realized in exactly solvable models such as the Toric Code or the Kitaev Honeycomb model. So the answer to whether anyons have creation and annihilation operators is yes.

However, as pointed out by @delete000, we need knowledge of the exclusion statistics to characterize the Fock space of an anyon type. In exactly solvable models this is apparent if there is a direct algebraic fractionalization as just described. But, I don't think there is a complete understanding of exclusion statistics for an anyon given set of fractional quantum numbers so I cannot completely answer part (1) of your question, although there is a recent discussion for the special case of parafermions.

**(2)** As pointed out in the comments, there are Toric Code models whose quasiparticles are extended operators that realize non-trivial mutual statistics. One good example is the recent exactly soluble 3D models by Lin and Levin realizing realizing braiding between points and loops.

The particle-loop braiding picture is also important for the gapless phase of the 3D variants of the Kitaev Honeycomb model (although gaplessness makes it difficult to identify the anyon in the wavefunction, it exists at the operator level) where the confinement transition takes place at finite temperature on the 3D lattice because the loops need not only to exist, but be very large to lead to the cancellation between spinon paths through about around the loops [cite: 1309.1171 and 1507.01639]. This is unlike the 2D case where a point defect already does the job, making the 2D Kitaev spin liquid unstable to finite temperatures.

For future convenience denote
$$
{\hat \phi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_j}
$$
$$
{\hat \chi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_{j+1}}
$$
The average you want to calculate reads then
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\langle 0 | \prod_n{{\hat \phi}_n} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} |0\rangle
$$
Starting with the regular CCR-s,
$$
\left[ {\hat c}_j, {\hat c}^\dagger_k \right]_\mp = \delta_{jk},\;\;\;\;\; \left[ {\hat c}_j, {\hat c}_k\right]_\mp = \left[ {\hat c}^\dagger_j, {\hat c}^\dagger_k\right]_\mp = 0
$$
obtain
$$
\left[ \sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j}, {\hat \phi}^\dagger_n\right] = {\hat \chi}^\dagger_n
$$
and
$$
{\hat \phi}^\dagger_n {\hat \chi}^\dagger_m = \pm {\hat \chi}^\dagger_m {\hat \phi}^\dagger_n
$$
$$
{\hat \phi}_n {\hat \chi}^\dagger_m = \pm {\hat \chi}^\dagger_m {\hat \phi}_n + \sum_j{A^*_{j+1, n}A_{j,m}}
$$
Now, in the average you need to calculate, use the above to successively move $\left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)$ past the orbital operators to its right:
$$
\left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} = {\hat \phi}^\dagger_1 \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>1}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} =
$$
$$
= \prod_{m=1}^{m=2} {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} + {\hat \phi}^\dagger_1 {\hat \chi}^\dagger_2 \prod_{n > 2} {{\hat \phi}^\dagger_n} =
$$
$$
= \prod_{m=1}^{m=2} {\hat \phi}^\dagger_m \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} \pm {\hat \chi}^\dagger_2 \prod_{n \neq 2} {{\hat \phi}^\dagger_n} = \dots =
$$
$$
= \prod_m {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) + \sum_m{ (\pm 1)^{m-1} {\hat \chi}^\dagger_m \prod_{n \neq m} {{\hat \phi}^\dagger_n}}
$$
The first term will annihilate the rhs vacuum, so only the sum remains. Now bring in the orbital operators on the left and flip each ${\hat \chi}^\dagger_m$ past them:
$$
\left(\prod_n{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m = \pm \left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m {\hat \phi}_1 + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} =
$$
$$
\left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m \prod_{l=1}^{l=2} {{\hat \phi}_l} + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} \pm \left(\prod_{n\neq 2}{{\hat \phi}_n}\right)\sum_j{A^*_{j+1, 2}A_{j,m}} = \dots =
$$
$$
= {\hat \chi}^\dagger_m \left(\prod_n{{\hat \phi}_n}\right) + \sum_l{(\pm 1)^{l-1}\left(\prod_{n\neq l}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, l}A_{j,m}} }
$$
The first term now annihilates the lhs vacuum, and after substituting everything the desired average becomes
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1} A^*_{j+1, l}A_{j,m} \langle 0 |\left(\prod_{n\neq l}{{\hat \phi}_n}\right)\left( \prod_{n' \neq m} {{\hat \phi}^\dagger_{n'}}\right) |0\rangle } =
$$
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1} A^*_{j+1, l}A_{j,m}\delta_{l,m} }
$$
and finally
$$
\langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, m}{ A^*_{j+1, m}A_{j,m} }
$$
There may be some bugs I missed, but this is the general idea.

## Best Answer

Let $\hat{a}^{\dagger}$ and $\hat{a}$ be the creation and annihilation operators respectively. Then what follows is a lifting from the usual bosonic algebra, viz:

Then \begin{align} \langle n|\hat{a}^{\dagger}|n \rangle \,&\propto\, \langle n | n+1 \rangle \\ &=0 \\ \langle n|\hat{a} \hat{a}^{\dagger}|n \rangle &= \langle n|(1 + \hat{a}^{\dagger}\hat{a})| n \rangle \\ &= \langle n|n \rangle + \langle n|\hat{a}^{\dagger}\hat{a})| n \rangle \\&= 1 + n \end{align} One can use the commutation relationd for these but it can get tedious, I prefer to utilise the ladder operator method.