= Solution
For the <Sheaf of relative Kähler differentials> on $X=\mathbb P^3_{\mathbb C}$, the cotangent form of the <Euler sequence> is
$$
0\longrightarrow\Omega_{X/\mathbb C}
\longrightarrow\mathcal O_X(-1)^{\oplus4}
\xrightarrow{(t_0,t_1,t_2,t_3)}\mathcal O_X\longrightarrow0.
$$
The last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed <Sheaf of relative Kähler differentials>. The <cohomology of twisting sheaves on projective space> gives $H^q(X,\mathcal O_X(-1))=0$ for every $q$, while $H^0(X,\mathcal O_X)=\mathbb C$ and its positive-degree groups vanish. The <long exact sequence in sheaf cohomology> therefore gives the <cotangent-sheaf cohomology of projective space> in this dimension:
$$
\boxed{H^q(X,\Omega_{X/W})\cong
\begin{cases}\mathbb C,&q=1,\\0,&q\ne1,\end{cases}
\qquad\chi(X,\Omega_{X/W})=-1.}
$$
The isomorphism for $q=1$ is the connecting map from the constant global sections of $\mathcal O_X$. The <Euler characteristic of a coherent sheaf> is the alternating sum of dimensions, explaining the minus sign.
For the curve, put $S=\mathbb C[t_0,t_1,t_2,t_3]$. It is a <unique factorization domain>, so irreducibility of $F$ makes $(F)$ a <prime ideal>. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the <integral domain> $S/(F)$. It follows that $(F,G)$ is a <regular sequence>. Its <Koszul complex> is an <exact sequence>, and graded sheafification gives
$$
0\longrightarrow\mathcal O_X(-12)
\xrightarrow{(-G,F)}\mathcal O_X(-5)\oplus\mathcal O_X(-7)
\xrightarrow{(F,G)}\mathcal O_X
\longrightarrow i_*\mathcal O_Z\longrightarrow0,
$$
where $i:Z\hookrightarrow X$ is the <closed immersion>. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.
Let $\mathcal I_Z$ be the <ideal sheaf of a closed subscheme>. Split this <Koszul resolution> into
$$
0\to\mathcal O_X(-12)\to\mathcal O_X(-5)\oplus\mathcal O_X(-7)
\to\mathcal I_Z\to0,
\qquad
0\to\mathcal I_Z\to\mathcal O_X\to i_*\mathcal O_Z\to0.
$$
For all integers $d$, $H^1(X,\mathcal O_X(d))=H^2(X,\mathcal O_X(d))=0$. Also $H^1(X,\mathcal O_X)=H^2(X,\mathcal O_X)=0$. The two <long exact sequences in sheaf cohomology>, together with <sheaf cohomology under a closed inclusion>, consequently identify
$$
H^1(Z,\mathcal O_Z)
\cong H^2(X,\mathcal I_Z)
\cong\ker\bigl(H^3(X,\mathcal O_X(-12))
\longrightarrow H^3(X,\mathcal O_X(-5))\oplus H^3(X,\mathcal O_X(-7))\bigr).
$$
By <Serre duality>, $H^3(X,\mathcal O_X(-m))\cong S_{m-4}^*$ for $m\geq4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\binom{d+3}{3}$. Thus the source has dimension $\binom{11}{3}=165$, and the two target spaces have dimensions $\binom{4}{3}=4$ and $\binom{6}{3}=20$. The <rank-nullity theorem> proves the requested bound:
$$
\boxed{\dim_{\mathbb C}H^1(Z,\mathcal O_Z)\geq165-4-20=141.}
$$
In fact equality holds. Under the <Serre duality> pairings, the dual of this map is
$$
S_1\oplus S_3\longrightarrow S_8,\qquad (a,b)\longmapsto-Ga+Fb.
$$
If $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. \b[Thus the lower bound is attained for every pair allowed in the question.] No smoothness assumption is needed. The <regular sequence> cuts out a <projective complete intersection> of dimension one. Also $H^0(X,\mathcal I_Z)=H^1(X,\mathcal I_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,\mathcal O_Z)=\mathbb C$. This is the <genus of a complete-intersection space curve>: the first cohomology has dimension $141$, equal to its <arithmetic genus>.
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