I am not absolutely sure about my answer and I would like to verify it please:
Let $K$ be an algebric closed field and let $I\subseteq K[x,y,z]$ be an ideal generated by $$ \begin{align} &f_1(x,y,z)=(x^2+y^2)^2,&&f_2(x,y,z)=(y^2+z^2)^3 \\&f_3(x,y,z)=(x^2+z^2)^5,&&f_4(x,y,z)=x+y+z+1 \end{align}$$ Describe explicitly $V(I)\subseteq K^3$. Find a necessity and sufficiency condition regarding to the field $K$ so that $V(I)\ne\emptyset$. Find generators to $\sqrt I$.
If $p=(x,y,z)\in V(I)$ then $$\begin{cases} 0=x^2+y^2\\0=x^2+z^2\\0=y^2+z^2 \end{cases} \Rightarrow\begin{cases} z^2=y^2\\z^2=-y^2 \end{cases}\Rightarrow2z^2=0$$
If $\operatorname{char}(K)\ne 2$ then $z=0\Rightarrow x=y=z=0\Rightarrow 0=1$ and thus $V(I)=\emptyset$.
If $\operatorname{char}(K)=2$ then $0=x^2+y^2=(x+y)^2\Rightarrow x=-y=y$. And in the same way $x=y=z$. Hence $1=-1=x+y+z=x+x+x=x=y=z$. Thus $V(I)=\{(1,1,1)\}$ and $V(I)\ne\emptyset\iff \operatorname{char}(K)=2$.
Generators of $\sqrt I$:
If $char(K)=2$ then the generators are $X+y,x+z,y+z$ and $1+x+y+z$.
If $\operatorname{char}(K) \ne 2$ then the generators are $f_1,f_2,f_3$ and $f_4$.
I think your answer about the generators is correct, but probably not the best way. If $char(K)\neq 2$, you have shown that $V(I)=\emptyset$, which means $I=(1)$. If $char(K)=2$ then $V(I)=\{(1,1,1)\}$, so $(x-1,y-1,z-1)$ is the only maximal ideal that contains $I$. By Hilbert's Nullstellensatz, the intersection of maximal ideals that contain $I$ is the radical of $I$. Therefore $\sqrt{I}=(x-1,y-1,z-1)$.