Why don't you simplify those fractions?

$\displaystyle \sum_{k=1}^{145} {\left[ \sin{\left(\frac{2\pi k}{8}\right)}-i\cos { \left( \frac {2\pi k}{14}\right)}\right] } = \sum_{k=1}^{145} {\sin{\left(\frac{k\pi}{4}\right)}}- i\sum_{k=1}^{145} {\cos { \left( \frac {k\pi}{7}\right)}}$

We note that $\displaystyle \sum_{k=1}^{145} {\sin{\left(\frac{2\pi k}{8}\right)}}$ and $\displaystyle \sum_{k=1}^{145} {\cos { \left( \frac {2\pi k}{14}\right)}}$ are cyclical and are equal to:

$\displaystyle \sum_{k=1}^{145} {\sin{\left(\frac{2\pi k}{8}\right)}} = \begin{cases} 0 & \text{when } n \text{ mod } 8 = 0 \\ \displaystyle \sum_{k=1}^{n \text{ mod }8} {\sin{\left(\frac{2\pi k}{8}\right)}} & \text{when } n \text{ mod } 8 \ne 0 \end{cases}$

$\displaystyle \sum_{k=1}^{145} {\cos{\left(\frac{2\pi k}{14}\right)}} = \begin{cases} 0 & \text{when } n \text{ mod } 14 = 0 \\ \displaystyle \sum_{k=1}^{n \text{ mod }14} {\cos{\left(\frac{2 \pi k}{14}\right)}} & \text{when } n \text{ mod } 14 \ne 0 \end{cases}$

For $n=145\quad \Rightarrow 145 \text{ mod }8 = 1 \quad \Rightarrow 145 \text{ mod }14 = 5$ , therefore,

$\displaystyle \sum_{k=1}^{145} {\left[ \sin{\left(\frac{2\pi k}{8}\right)}-i\cos { \left( \frac {2\pi k}{14}\right)}\right] } = \sum_{k=1}^{1} {\sin{\left(\frac{k\pi}{4}\right)}}- i\sum_{k=1}^{5} {\cos { \left( \frac {k\pi}{7}\right)}} \\ = \sin{\dfrac{\pi}{4}} -i \left( \cos {\dfrac {\pi}{7}} + \cos {\dfrac {2\pi}{7}} + \cos {\dfrac {3\pi}{7}} + \cos {\dfrac {4\pi}{7}} +\cos {\dfrac {5\pi}{7}} \right) \\ = \sin{\dfrac{\pi}{4}} -i \left( \cos {\dfrac {\pi}{7}} + \cos {\dfrac {2\pi}{7}} + \cos {\dfrac {3\pi}{7}} - \cos {\dfrac {3\pi}{7}} -\cos {\dfrac {2\pi}{7}} \right) \\ = \sin{\dfrac{\pi}{4}} -i\cos {\dfrac {\pi}{7}} = \dfrac{1}{\sqrt{2}} -i\cos {\dfrac {\pi}{7}}$

$\Rightarrow a + b = 2 + 7 = \boxed{9}$