use comparison test to show divergence or convergence I'm not sure if my reasoning is correct. 
a) $\displaystyle \sum_{n=2}^{\infty} \frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} = \sum_{n=2}^{\infty} a_n $
here I think it suffice to show that for big enough $n$ we have all logs $>1$ and $\sin n \ge -1$ so we have $\displaystyle \sum_{n=2}^{\infty}\frac{-9}{n} \le a_n $ so $\sum_{n=2}^{\infty} a_n$ diverges.
b) $\displaystyle \sum_{n=2}^{\infty} (\frac{n^2+3n+10}{n^2+5n+17})^{n^2(\sqrt{n+1}-\sqrt{n-1})}= \sum_{n=2}^{\infty} b_n$ 
we have $\displaystyle \sum_{n=2}^{\infty} b_n= \sum_{n=2}^{\infty}(1-\frac{2n+7}{n^2+5n+17})^{\frac{2n^2}{\sqrt{n+1}+\sqrt{n-1}}}$ and here I don't know how to compare it

Part (b):
The exponent of $b_n$ satisfies
$$\frac{2n^2}{\sqrt{n+1}+\sqrt{n-1}}< \frac{2n^2}{\sqrt{n+1}}< 2n\sqrt{n}.$$
For sufficiently large $n$
$$1-\frac{2n+7}{n^2+5n+17}= 1-\frac{2}{n}\frac{1+7/(2n)}{1+5/n+17/n^2}< 1 - \frac{1}{n},$$
using
$$\lim_{n \to \infty}\frac{1+7/(2n)}{1+5/n+17/n^2} = 1 \implies\frac{1+7/(2n)}{1+5/n+17/n^2} > \frac1{2}.$$
Using the inequality $\displaystyle (1+x/n)^{n+1}> e^x,$ we have
$$\left(1 - \frac{1}{n}\right)^n= \left(1 + \frac{1}{n-1}\right)^{-n}<e^{-1}.$$
Hence, for sufficiently large n,
$$b_n < \left(1 - \frac{1}{n}\right)^{2n\sqrt{n}}< e^{-2\sqrt{n}}.$$
Note that $e^n$ grows faster than any power of $n$.  We easily can find a dominating convergent p-series.
$$e^{2\sqrt{n}} = \sum_{k=0}^{\infty}\frac{(2\sqrt{n})^k}{k!}> \frac{(2\sqrt{n})^4}{4!}=\frac{2n^2}{3}\\ \implies e^{-2\sqrt{n}} < \frac{3}{2}n^{-2}.$$
Therefore, by the comparison test
$$\sum_{n=2}^{\infty} b_n < \sum_{n=2}^{\infty} e^{-2\sqrt{n}}< \frac{3}{2}\sum_{n=2}^{\infty} n^{-2} = \frac{3}{2}\left(\frac{\pi^2}{6}-1\right).$$
Part (a):
Note that for sufficiently large $n$
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > \frac{\ln^5{(2n^7+13)-10}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > 1.$$
Also 
$$ \lim_{n \to \infty}\frac{(n^{\frac{7}{8}}+2\sqrt{n}-1)}{n} = 0, \\ n+ (-1)^n \leqslant n+1,$$
and for sufficiently large $n$ we have
$$ \ln(n^{\frac{7}{8}}+2\sqrt{n}-1) < \ln n, \\ \ln \ln (n+ (-1)^n) \leqslant \ln \ln (n+1),$$
Thus,
$$a_n = \frac{\ln^5{(2n^7+13)+10\sin(n)}}{n\cdot \ln^6{(n^\frac{7}{8}}+2\sqrt{n}-1)\cdot\ln{\ln{(n+(-1)^n}})} > \frac{1}{n \cdot \ln n \cdot \ln \ln (n+1)} \\ > \frac{1}{(n+1) \cdot \ln (n+1) \cdot \ln \ln (n+1)} .$$
The series with terms given by the RS of the above inequality diverges by the integral test, since
$$\int_2^{\infty}\frac{dx}{(x+1) \cdot \ln (x+1) \cdot \ln \ln (x+1) }= \lim_{x \to \infty}[\ln \ln \ln(x+1)- \ln \ln \ln (3)] = \infty.$$
