Some quantum mechanics books tell us that the definitions of hermitian are
If $\langle\psi|A\phi\rangle=\langle B\psi|\phi\rangle$ for linear operators, then $B=A^\dagger$
If $\langle\psi|C\phi\rangle=\langle D\psi|\phi\rangle^*$ for anti-linear operators, then $D=C^\dagger$
Why are defintions different between linear operators and anti-linear operators?
Let $\lambda$ be any complex scalar. Then $$\langle \psi|C(\lambda \phi)\rangle = \langle\psi|\overline \lambda C\phi\rangle = \overline \lambda\langle\psi|C\phi\rangle$$
by the definition of anti-linearity. If we use the definition we normally use for linear operators, then we would find $$\langle \psi | C(\lambda \phi)\rangle = \langle D\psi|\lambda \phi\rangle$$ but also that $$\overline \lambda \langle\psi|C\phi\rangle = \overline \lambda\langle D \psi|\phi\rangle = \langle D\psi|\overline \lambda \phi\rangle$$
which is inconsistent. Using the slightly modified definition fixes this problem.
Are they equivalent to the definition $Q^\dagger=(Q^T)^∗$ in linear algebra, where $Q^T$ is the transpose of $Q$?
The first one is, yes.
If you imagine two arbitrary column vectors $\mathbf x$ and $\mathbf y$ and matrices $A$ and $B$, then the first definition becomes $$\mathbf x^\dagger A \mathbf y = (B\mathbf x)^\dagger \mathbf y \implies B=A^\dagger$$
where the dagger denotes conjugate transposition. But this should be clear, since $$(B\mathbf x)^\dagger = \mathbf x^\dagger B^\dagger$$ so $$\mathbf x^\dagger A \mathbf y= \mathbf x^\dagger B^\dagger \mathbf y$$ If that equality holds for all $\mathbf x$ and $\mathbf y$, then $A=B^\dagger \iff A^\dagger = B$.
The case of anti-linear operators is a bit more subtle because an anti-linear operator cannot be written as a matrix all by itself; it has to be a matrix plus a complex conjugation, and therefore cannot be realized if we restrict ourselves to standard matrix algebra.
