Given the ring $Z_5[x]$, the ideal $I=(x^2+3)$ and the polynomial $f(x)=14x^2+k$, for which $k \in \{0,1,2,3,4 \} $ it holds true that f(t)=1 in $R/I$? (where $t$ is the class of $x$ in $R/I$)
I am not sure how to calcualte $f(t)$ in $R/I$. I thought that $f(t)=14t^2+k=4t^2+k$ and because the Euclidian division gives $$4x^2+k=4(x^2+3)+ (-12+k)$$ it has to be $$f(t)=-12+k$$ so then $$f(t)=1 \leftrightarrow -12+k=1 \leftrightarrow k=3.$$
Can you check my work?
You've arrived at the right answer! Here are some additional thoughts and details that might be useful.
Since $14 = 4$ in $\mathbb{Z}_5$, let's write (as you've done) $f(x) = 4x^2 + k$. Now, if $f(x)$ reduces to $1$ in $R/I$, then there is a polynomial $g(x) \in \mathbb{Z}_5[x]$ such that $f(x) = (x^2 + 3)g(x) + 1$. (This is just the definition of congruence modulo an ideal.) Since $f(x)$ has degree 2, and since
${\rm deg}((x^2 + 3)g(x) + 1) = {\rm deg}(x^2 + 3) + {\rm deg}(g(x)) = 2 + {\rm deg}(g(x))$,
it follows that ${\rm deg}(g(x)) = 0$. In other words, $g(x)$ must be constant, say, $g(x) = c$. This gives $4x^2 + k = cx^2 + 3c + 1$, and since two polynomials are equal iff they have the same coefficients, it follows that $c = 4$. Since $3c = 12$ is congruent to $2$ mod $5$, we have $f(x) = 4x^2 + 2 + 1 = 4x^2 + 3$. Finally, it follows that $k = 3$.