The first calculator is doing degrees. $\cos 35^\circ \approx 0.819$. This is reasonable because $0< 35 < 90$ so the $\sin/\cos$ values will both be positive. $30 < 35 < 45$ so $\cos 30^\circ =\frac {\sqrt 3}2 > \cos 35^\circ > \cos 45^\circ =\frac 1{\sqrt 2}$ so the value $0.819...$ is reasonable. $-0.9036...$ is not.
The second calculator is doing radians and for radian input $35$ is a strange input but not impossible. $35 = \frac {35}{\pi} \pi \approx 11.14 \pi$. As trig values repeat every $2\pi$ turns, $\cos 35 = \cos 11.14 \pi = \cos 1.14\pi$. As $\cos \pi = -1$, we have $\cos 1.14$ will be "kind of" close to $-1$. And $-0.9036...$ is reasonable for radians.
Alternatively. $360^\circ = 2\pi \ radians$ so $1\ radian = \frac {360}{2\pi}^\circ$ and $35\ radians = 35\times \frac {360}{2\pi}^\circ \approx 2005.35^\circ$. As trig values repeat every $360^\circ$, we know $\cos 2005.35^\circ = \cos (5\cdot 360 + 205.35)^\circ = \cos 205.35^\circ = \cos (180^\circ + 25.35^\circ)$. As $\cos 180 = -1$ then $\cos 205.35$ should be "kind of" close to $-1$ and that's a reasonable answer for $\cos 2005.35^\circ$.
(More guestimation: $\cos (180 + k)^\circ = - \cos k^\circ$ so $\cos 205^\circ = \cos (180 + 25)^\circ = -\cos 25^\circ$. As $0 < 25 < 30$ we have $1=\cos 0 > \cos 25 > \cos 30 = \frac {\sqrt 3}2 \approx 0.866$ so $\cos 25^\circ \approx 0.9036...$ is a reasonable answer and $-0.9306...$ is a reasonable answer for $\cos 2005^\circ$.)
tl;dr: Doing trig with degrees (1 circle = $360^\circ$) and doing trig with radians (1 circle = $2\pi$ radians) will have very different results. One calculator is using degrees. The other is using radians.... That's my final answer.