You did both correctly.
For the first part, you can figure out most if it using only geometry. A right triangle with two equal sides is a 45-45-90 triangle. A right triangle with sides $\displaystyle x, x\sqrt{3}$ is a 30-60-90 triangle (where $\displaystyle \theta_2$ is the 30 degree angle). So, $\displaystyle \theta_1 = \dfrac{\pi}{4}, \theta_2 = \dfrac{\pi}{6}$. This means $\displaystyle \theta = \pi - \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{7\pi}{12}$.
You get the same answer either way.
I was taught that the sides of a 30-60-90 triangle will have lengths $\displaystyle x, x\sqrt{3}, 2x$ where $\displaystyle x$ is the length of the side opposite the 30 degree angle, $\displaystyle x\sqrt{3}$ is the length of the side opposite the 60 degree angle, and $\displaystyle 2x$ is the length of the hypotenuse in geometry class. I assumed it was still taught. As for the other angle, convert degrees to radians: $\displaystyle 45\dfrac{\pi}{180} = \dfrac{\pi}{4}$.
I got a different result when applying the half-angle formula. I got:
$\displaystyle \sin\left(\dfrac{7\pi}{12}\right) = \sin\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{ \pi}{12}\right) + \sin\left(\dfrac{\pi}{12}\right) \cos\left(\dfrac{\pi}{2}\right) = \cos\left(\dfrac{\pi}{12}\right)$.
Since $\displaystyle \cos(2\theta) = 2\cos^2\theta-1$, the half-angle formula gives:
$\displaystyle \cos \theta = \sqrt{\dfrac{\cos(2\theta)+1}{2}}$
Plugging in $\displaystyle \theta = \dfrac{\pi}{12}$ you have
$\displaystyle \cos \left( \dfrac{\pi}{12}\right) = \sqrt{\dfrac{\cos\left(\dfrac{\pi}{6}\right)+1}{2} } = \sqrt{\dfrac{\dfrac{\sqrt{3}}{2}+1}{2}} = \dfrac{1}{2}\sqrt{\sqrt{3}+2}$
So, I got a different result.
Edit: I just checked wolframalpha. Apparently, they are the same.
I calculated mine using vector product: $\displaystyle \sin\theta=\frac{|u\times v|}{|u|\cdot|v|}$. It's also possible to use dot product to find cosine first. If the angles are arbitrary, I would guess the problem authors' intention was to use the formula
$\displaystyle \sin(\pi-\theta_1-\theta_2)= \sin(\theta_1+\theta_2)=\sin(\theta_1)\cos(\theta_ 2)+ \cos(\theta_1)\sin(\theta_2)$
The values in the right-hand side can be found by definition.