= Solution
A <formal power series> $f=\sum_{n\geq0}a_nx^n$ is a <unit> exactly when its constant coefficient is a <unit> in $A$. If $fg=1$, constant coefficients give $a_0b_0=1$. Conversely, for invertible $a_0$, set
$$
b_0=a_0^{-1},\qquad b_n=-a_0^{-1}\sum_{i=1}^na_ib_{n-i}\quad(n\geq1).
$$
Then every positive-degree coefficient of $fg$ is zero and its constant coefficient is one, so $g=\sum b_nx^n$ is an inverse. This is a formal coefficient calculation and requires no convergence.
Since $A$ is a <local ring>, its nonunits are exactly $\mathfrak m$. The nonunits of $B$ are consequently the series with $a_0\in\mathfrak m$, which are exactly $I=\mathfrak mB+xB$: such a series is $a_0+xg$, and conversely an element of $I$ has constant coefficient in $\mathfrak m$. The map $B\to k$ taking the constant coefficient modulo $\mathfrak m$ is <surjective> with <kernel> $I$, so $I$ is a <maximal ideal>. Every proper <ideal> consists of nonunits and hence lies in $I$, proving that $I$ is the unique <maximal ideal>. Thus
$$
\boxed{B\text{ is local},\qquad\operatorname{Max}(B)=\{I\},\qquad B/I\cong k.}
$$
This proves the description of <formal power series over a local ring>.
To compute the <cotangent space of a formal power series ring>, expand
$$
I^2=\mathfrak m^2B+x\mathfrak mB+x^2B.
$$
Define the $k$-linear map
$$
\Theta:(\mathfrak m/\mathfrak m^2)\oplus k\longrightarrow I/I^2,\qquad(\bar a,\bar b)\longmapsto a+bx\pmod{I^2}.
$$
Changing a lift $a$ by $\mathfrak m^2$ or a lift $b$ by $\mathfrak m$ changes the image by $I^2$, so the map is well-defined. Every series in $I$ is congruent modulo $x^2B\subseteq I^2$ to its first two coefficients, proving surjectivity. If $a+bx\in I^2$, its constant coefficient lies in $\mathfrak m^2$ and its linear coefficient lies in $\mathfrak m$, so both input classes are zero. This proves injectivity. Therefore
$$
\boxed{I/I^2\cong(\mathfrak m/\mathfrak m^2)\oplus k,\qquad\dim_k(I/I^2)=d+1.}
$$
Finally, part (a) gives generators $a_1,\ldots,a_d$ of $\mathfrak m$, so $I=(a_1,\ldots,a_d,x)$. Use the <Krull height theorem> in its lecture form: in a <Noetherian ring>, a prime minimal over an <ideal> generated by $r$ elements has height at most $r$. The problem allows Noetherianity of $B$ to be assumed. Apply the theorem to $I$ itself, whose height is the <Krull dimension> of the <local ring> $B$. Hence $\boxed{\dim B=\operatorname{ht}(I)\leq d+1}$.
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