The answer to the first question is yes.
Let $V$ be a finite dimensional complex vector space, and let $\langle\cdot,\cdot\rangle:V\times V\rightarrow\Bbb{C}$ be an inner product. Fix a basis $\{e_1,\ldots,e_n\}$, so that for any two vectors $u,v\in V$, we have
$ \langle u,v\rangle=\left\langle\sum_{j=1}^n\alpha_je_j,\sum_{k=1}^n\beta_ke_k\right\rangle=\sum_{j=1}^n\sum_{k=1}^n\alpha_j\bar{\beta}_j\langle e_j,e_k\rangle $
So if we define $M=(M_{ij})_{n\times n}=(\langle e_i,e_j\rangle)_{n\times n}$, we have exactly the expression
$ \langle u,v\rangle = u^\intercal M \bar{v} $
as desired. It is easy to check that $M$ is Hermitian - this follows from conjugate symmetry of the inner product: $\overline{\langle x,y\rangle}=\langle y,x\rangle$. Positive definiteness follows from the positivity of the inner product - $\langle x,x\rangle>0$ for all $x\in V\backslash\{0\}$.
For the second claim, it suffices to show the identity $x^\intercal M \bar{y}=y^*\overline{M} x$. This is straightforward:
$ x^\intercal M\bar{y}=\sum_j\alpha_j\sum_kM_{jk}\bar{\beta}_k=\sum_k\bar{\beta}_k\sum_jM_{jk}\alpha_j=y^*\overline{M}x $
Notice of course that we must then use the matrix $\overline{M}$, i.e. the same matrix won't work. (Thanks to @user1551 for pointing this out)