Using complex numbers, can sine be greater than $1?$

Yes.

For a complex number $z$, we have \begin{align} \sin z &= \frac{1}{2i}(\exp(iz) - \exp(-iz)) \\ \cos z &= \frac{1}{2}(\exp(iz) + \exp(-iz)) \end{align} so in your problem, we have \begin{align} 10 &= \frac{1}{2i}(\exp(iz) - \exp(-iz)) \\ 10i &= \frac{1}{2}(\exp(iz) - \exp(-iz)) & \text{mult through by $i$}\\ i \sqrt{99} &= \frac{1}{2}(\exp(iz) + \exp(-iz)) \end{align} Adding the last two equations gives us $$ i(10 + \sqrt{99} = \exp{iz} $$ so $$ \log i(10 + \sqrt{99} ) = iz $$ and hence $$ z = \frac{\log (i(10 + \sqrt{99}) ) }{i}. $$

Here "log" should be taken as the principal value of the natural logarithm (you may need to look this up!), and there's a second answer where we set the cosine to be $-i\sqrt{99}$, but I didn't bother writing it out.

Yes, it can.

Let us find such $z$ so that $\sin(z) = 10$.

Firstly, as a consequence of the Euler's formula: $ e^{iz} = \cos(z) + i \cdot \sin(z)$,

we have the following two identities which hold for all complex numbers $z$:

$$ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \\ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} $$

Then:

$$ \sin(z) = 10 \\ \frac{e^{iz} - e^{-iz}}{2i} = 10 \\ e^{iz} - e^{-iz} - 20i = 0 \\ (e^{iz})^2 - 20i\cdot(e^{iz}) - 1 = 0 \\ $$

Solving the quadratic equation:

$$ \sqrt{D}=\sqrt{(-20i)^2 - 4 \cdot 1 \cdot (-1)} = \sqrt{-396} = 6i\sqrt{11} \\ (e^{iz}) = \frac{-(-20i) \pm \sqrt{D}}{2} = i \cdot (10 \pm 3\sqrt{11}) \\ $$

Complex logarithm:

$$ iz = Ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) = \ln\left(i \cdot \left(10 \pm 3\sqrt{11}\right)\right) + i \cdot 2\pi k, \qquad k \in \mathbb{Z}\\ iz = \ln(i) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ iz = \ln\left(e^{i\frac{\pi}{2}}\right) + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ iz = i\frac{\pi}{2} + \ln\left(10 \pm 3\sqrt{11}\right) + 2\pi ki \\ $$

The result:

$$ z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(10 \pm 3\sqrt{11}\right), \qquad k \in \mathbb{Z} $$

$\cos(z) = \sqrt{-99}$ can be solved in exactly the same way, giving the result:

$$ z = \frac{\pi}{2} + 2\pi k - i \cdot \ln\left(3\sqrt{11} \pm 10\right), \qquad k \in \mathbb{Z} $$

The idea is taken from this video: https://www.youtube.com/watch?v=3C_XD_cCeeI