= Solution
The three answers, in the given order, are \b[no, yes, yes].
For $\mathbb Z[x]$, consider the <ideal> $I=(2,x)$. Its members have even constant coefficient, so $1\notin I$ and the ideal is proper. If $I=(g)$, then $g$ divides the constant $2$, forcing $g$ to be constant by additivity of <polynomial degree> in this <integral domain>. Since $g$ also divides $x$, it must divide the coefficient $1$, giving $g=\pm1$. That would make $I$ the whole ring, a contradiction. Thus $\mathbb Z[x]$ is not a <principal ideal domain>, and part (a) shows it cannot be a <Euclidean domain>. This is the <nonprincipal ideal generated by a prime and an indeterminate> obstruction.
Evaluation at $i$ gives the <ring isomorphism>
$$
\boxed{\mathbb Z[x]/(x^2+1)\cong\mathbb Z[i].}
$$
Indeed, division by the <monic polynomial> $x^2+1$ leaves a unique remainder $a+bx$, and its value $a+bi$ is zero only when both integer coefficients vanish. Thus the quotient maps bijectively, with operations preserved, onto the <Gaussian integers>. These form a <Euclidean domain> for the <Gaussian integer norm> $N(a+bi)=a^2+b^2$, a standard Euclidean example allowed in this part. Geometrically, rounding the real and imaginary coordinates gives a quotient error with squared modulus at most $1/2<1$.
Finally,
$$
\boxed{\mathbb Z[x,y]/(2,x)\cong\mathbb F_2[y],}
$$
via $f(x,y)\mapsto f(0,y)\bmod2$. Every polynomial over $\mathbb F_2$ occurs. The kernel consists exactly of polynomials whose $x$-independent part has even coefficients: write $f(x,y)=xg(x,y)+h(y)$ to see that this is precisely $(2,x)$. Hence the induced quotient map is a bijective <ring homomorphism>. Its target is a <Euclidean domain> with Euclidean function $\deg$: a <polynomial ring over a field is Euclidean> by the division algorithm in part (a).
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