= Solution
Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi)=\sum_{a\bmod N}\chi(a)e^{2\pi ia/N}$ for its <Gauss sum of a Dirichlet character>. We use the analytic branch of $\sqrt{-i\tau}$ that is positive when $\tau$ is positive imaginary. With the convention $\widehat F(t)=\int_{\mathbb R}F(x)e^{-2\pi ixt}\,dx$, the <Gaussian Fourier transform> and its differentiated version are
$$
\widehat{e^{\pi i\tau x^2}}(t)=(-i\tau)^{-1/2}e^{-\pi it^2/\tau},\qquad
\widehat{x e^{\pi i\tau x^2}}(t)=\frac{t}{\tau}(-i\tau)^{-1/2}e^{-\pi it^2/\tau}.
$$
The first formula follows by completing the square, first on the imaginary axis and then by <analytic continuation>; the second follows by differentiating the <Fourier transform> in $t$. All the functions are <Schwartz functions> on the real axis.
For completeness, the <finite Fourier transform of a primitive Dirichlet character> satisfies
$$
\sum_{a\bmod N}\chi(a)e^{2\pi ira/N}=g(\chi)\overline\chi(r).
$$
For a unit $r$ this is substitution by its inverse. If $p\mid(r,N)$, primitivity supplies a unit $u\equiv1\pmod{N/p}$ with $\chi(u)\ne1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi(u)$, so the sum vanishes. This proves the identity also on nonunits. Finite <Parseval identity> now gives $|g(\chi)|^2=N$, since exactly $\varphi(N)$ frequencies have nonzero character values. In particular this <Gauss sum of a Dirichlet character> is nonzero.
Apply <Poisson summation> to each residue class: $\sum_jF(a+Nj)=N^{-1}\sum_r e^{2\pi ira/N}\widehat F(r/N)$. Multiply by $\chi(a)$ and sum over $a$. The formulas above give the two <character theta transformations>
$$
\boxed{\theta_\chi(\tau)=\frac{g(\chi)}{N\sqrt{-i\tau}}\,
\theta_{\overline\chi}\!\left(-\frac1{N^2\tau}\right)\quad\text{if }\chi(-1)=1,}
$$
and
$$
\boxed{\widetilde\theta_\chi(\tau)=\frac{g(\chi)}{iN^2(-i\tau)^{3/2}}\,
\widetilde\theta_{\overline\chi}\!\left(-\frac1{N^2\tau}\right)\quad\text{if }\chi(-1)=-1.}
$$
In the odd case the prefactor originally reads $g(\chi)/(N^2\tau\sqrt{-i\tau})$; using $\tau=i(-i\tau)$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd <Dirichlet characters>.
To deduce the products from these transformations, use two specific <primitive Dirichlet characters>. The even character $\chi_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its <Gauss sum of a Dirichlet character> is $2\sqrt3=\sqrt{12}$. Put
$$
A(\tau)=\tfrac12\theta_{\chi_{12}}(\tau/12)=\tfrac12\sum_n\chi_{12}(n)q^{n^2/24}.
$$
The even transformation gives $A(-1/\tau)=\sqrt{-i\tau}\,A(\tau)$; also $A(\tau+1)=e^{\pi i/12}A(\tau)$, since each nonzero term has $n^2\equiv1\pmod{24}$. Thus $A^{24}$ transforms as a weight-twelve <modular form> under the generators $S,T$ of the <modular group>. It is holomorphic and starts with $q$, so it is a normalized <cusp form>. The proof in Question 1 shows $S_{12}=\mathbb C\Delta_0$, giving $A^{24}=\Delta_0$. In particular $A$ has no interior zeros, by the <valence formula for the modular group>.
The <Serre derivative> argument in Question 2 gives $D\Delta_0=E_2\Delta_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau)=q^{1/24}\prod_{n\geq1}(1-q^n)$ is nonzero and has the same <logarithmic derivative>, since $DP/P=1/24-\sum\sigma_1(n)q^n$. Therefore $A/P$ is constant on the connected <complex upper half-plane>; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since $\chi_{12}(6r+1)=(-1)^r$, we get
$$
A=q^{1/24}\sum_{r\in\mathbb Z}(-1)^r q^{r(3r+1)/2},
\qquad
\boxed{\prod_{n\geq1}(1-q^n)=\sum_{r\in\mathbb Z}(-1)^r q^{r(3r+1)/2}.}
$$
This is the <Euler pentagonal identity>, derived by the <character theta proof of the eta product>.
For the second product use the odd primitive character $\chi_4(1)=1$, $\chi_4(3)=-1$, with <Gauss sum of a Dirichlet character> $2i$. Set
$$
B(\tau)=\tfrac12\widetilde\theta_{\chi_4}(\tau/4)=\tfrac12\sum_n n\chi_4(n)q^{n^2/8}.
$$
The odd transformation gives $B(-1/\tau)=(-i\tau)^{3/2}B(\tau)$ and the series gives $B(\tau+1)=e^{\pi i/4}B(\tau)$. Thus $B^8$ is another normalized weight-twelve <cusp form>, and $B^8=\Delta_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now gives
$$
B=q^{1/8}\sum_{r\geq0}(-1)^r(2r+1)q^{r(r+1)/2}.
$$
Since $B=A^3=q^{1/8}\prod_{n\geq1}(1-q^n)^3$, the <Jacobi cubic product identity> follows:
$$
\boxed{\prod_{n\geq1}(1-q^n)^3=\sum_{r\geq0}(-1)^r(2r+1)q^{r(r+1)/2}.}
$$
All pairings and differentiations take place in locally uniformly convergent <theta functions> or products on $|q|<1$.
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