Evaluate $\oint_\gamma \frac{e^z}{z+5}\,{\rm d}z$ where
$\gamma(t)=1+2e^{it},\,0\leq t\leq 2\pi$.
$\gamma(t)=1+7e^{it},\,0\leq t\leq 2\pi$.
My method is by using Cauchy's integral formula, and let $f(z)=e^z$, then
$$\oint_\gamma \frac{e^z}{z+5}\,{\rm d}z = 2\pi if(-5)=2\pi i e^{-5}$$ What's the difference between 1. and 2.?
Your answer for the second part is correct.
On the other hand, the circle $|z+1|=2$ given by the loop in part 1 does not contain -5. Therefore, CIF is non-applicable and the contour can be deformed to any point besides the problem at -5 (aka a pole of order 1 or simple pole), giving $0$.
On the other hand, the loop in part 2 is $|z+1|=7$, which contains -5. This forces you to use CIF.