Let's start with the first equation, which is, incidentally, a good example to show that, unlike the sine of a real number, the sine of a complex number can be larger than 1 in its norm.
On setting \(x:=\text{Re }z\in\mathbb R\) and \(y:=\text{Im }z\in\mathbb R\), it is straightforward to get the following system of equations: \begin{cases}\cos x \left(e^{-y}-e^{y}\right)=0 \\ \sin x = \frac{4}{e^y+e^{-y}}, \end{cases} the second equation having solutions for all \(y\) such that \(\frac{4}{e^y+e^{-y}} \le 1\). The first equation is satisfied if either \(\cos x = 0\) or \(y=0\). If \(y=0\), then the second equation has no solution. If \(\cos x = 0\), i.e. \(x=\frac{\pi}{2}+n\pi,\) \(n\in \mathbb Z\), the second equation becomes, on setting \(\alpha:=e^y\), \[ (-1)^n \alpha^2 - 4\alpha + (-1)^n=0,\]
whose solutions have the form \(\alpha_{\pm} := (-1)^n (2\pm\sqrt 3)\). Hence, we have to choose \(n\) even, for satisfying the requirement that \(\alpha_{\pm}\) be nonnegative, and so we have \(e^{y_{\pm}} = 2\pm\sqrt 3\Rightarrow y_{\pm} = \ln(2\pm\sqrt 3)\). All in all, on setting \(n:=2k,\,k\in\mathbb Z\), we find that the \(z\in\mathbb C\) which satisfy \(\sin z = 2\) have the forms \(\begin{equation}\begin{split}z_+ := \frac{\pi}{2} + 2k\pi + i\,\ln(2+\sqrt 3),\\z_- := \frac{\pi}{2} + 2k\pi + i\,\ln(2-\sqrt 3),\end{split}\end{equation}\), with \(k\in\mathbb Z.\) Hope there are no mistakes...
A faster method could have been to note that, when \(\cos x=0\), we must have \(\sin x = \pm 1\) (in this case, \(\sin x = 1\) since the right-hand side of the second equation is nonnegative). This holds if and only if \(e^y+e^{-y}=4\), which can be rapidly solved with respect to \(y\). Condition \(\sin x = 1\) also implies that we must replace \(n\) by an even number \(2k\).
(How can I centre a system of equations? \begin{equation} doesn't work...)