The first one is: consider the maximal ideal $\mathfrak{m} = ([x^2],[x^3]) $ of the local ring $A=k[x^2,x^3]/(x^4)$. Is it true that $\mathfrak{m}^2=0$? In my opinion, $\mathfrak{m}^2$ is generated by $[x^4],[x^5]$ and $[x^6]$. I can see why the first and the last are $0$, by why should $[x^5]=0$ in $A$, since $[x]$ is not in $A$?
Second question: consider the ring $A=k[x,y,z,w]/(xz-y^2,yz-xw,z^2-yw)$. I am told that its fraction field is $K=k\left(\frac{[y]}{[z]}\right)$, but I cannot see how one could recover for instance $[x] \in A \subseteq K$ as a $k-$multiple of some (possibly negative) power of $\frac{[y]}{[z]}$.
Note: throughout, the square brackets denote equivalence classes in the quotient ring (except in the obvious cases when they just adjoin variables to $k$).
You should use $k[x^2,x^3]/(x^4)\cong k[y,z]/(y^3-x^2,y^2)$, see this answer. So $\mathfrak m=(z,y)$ represents the origin and $\mathfrak m^2=(zy)$ is not the zero ideal, because $zy$ does not vanish on the entire cusp.
Furthermore, $R:=k[x,y,z,w]/(xz-y^2,yz-xw,z^2-yw)$ is a naturally graded ring and the elements of $k([y]/[z])$ all have degree zero. Hence, it cannot be the quotient field of $R$. It is, however, the degree $0$ component of $K$. Note that $$ K_0 = k\left( \tfrac{[x]}{[z]}, \tfrac{[y]}{[z]}, \tfrac{[w]}{[z]} \right) $$ and we have \begin{align*} \frac{[w]}{[z]}&=\frac{[xw]}{[xz]}=\frac{[yz]}{[y^2]}=\frac{[z]}{[y]}=\left(\frac{[y]}{[z]}\right)^{-1}, \\ \frac{[x]}{[z]}&=\frac{[xw]}{[zw]}=\frac{[yz]}{[zw]}=\frac{[y]}{[w]} =\frac{[y^2]}{[yw]}=\frac{[y^2]}{[z^2]}=\left(\frac{[y]}{[z]}\right)^2, \end{align*} which shows that $K_0=k\left( \tfrac{[y]}{[z]} \right)$.