#### Answer

$k(0)=3$
$k(2)=-5$
$k(-2)=3$
$k(\sqrt{2})=1-2\sqrt{2}$
$k(a+2)=-a^{2}-6a-5$
$k(-x)=-x^{2}+2x+3$
$k(x^{2})=-x^{4}-2x^{2}+3$

#### Work Step by Step

We are given:
$k(x)=-x^{2}-2x+3$
We evaluate:
$k(0)=-0^{2}-2(0)+3=0-0+3=3$
$k(2)=-2^{2}-2(2)+3=-4-4+3=-5$
$k(-2)=-(-2)^{2}-2(-2)+3=-4+4+3=3$
$k(\sqrt{2})=-(\sqrt{2})^{2}-2(\sqrt{2})+3=-2-2\sqrt{2}+3=1-2\sqrt{2}$
$k(a+2)=-(a+2)^{2}-2(a+2)+3=-(a^2+4a+4)-2a-4+3=-a^{2}-6a-5$
$k(-x)=-(-x)^{2}-2(-x)+3=-x^{2}+2x+3$
$k(x^{2})=-(x^{2})^{2}-2(x^{2})+3=-x^{4}-2x^{2}+3$