= Solution
The drift makes the natural measure weighted. Let
$$
\mathcal A=\partial_x^2+\beta\partial_x
=e^{-\beta x}\partial_x(e^{\beta x}\partial_x),\qquad
w(x)=e^{\beta x}.
$$
Under homogeneous <Neumann boundary conditions>, the <differential operator> is <self-adjoint> in the <weighted inner product> $\langle u,v\rangle_w=\int_0^Lw\overline u v\,dx$. Indeed <integration by parts> gives
$$
\langle u,\mathcal Av\rangle_w
=-\int_0^L w\overline{u'}v'\,dx
$$
when the derivative boundary terms vanish. In particular its <eigenvalues> are nonpositive and it has the constant stationary <eigenfunction>. We first construct the <weighted Neumann heat kernel with constant drift>, then incorporate the two prescribed coordinate derivatives.
Put $b=\beta/2$. The <Robin gauge transform for constant drift>, $\phi=e^{-bx}\psi$, turns $\mathcal A\phi=-\lambda\phi$ into
$$
\psi''+(\lambda-b^2)\psi=0,\qquad
\psi'(0)=b\psi(0),\quad\psi'(L)=b\psi(L).
$$
These are coordinate-derivative <Robin boundary conditions>, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi=\cos(kx)+(b/k)\sin(kx)$. The right condition then reduces to $(k+b^2/k)\sin(kL)=0$, giving $k_n=n\pi/L$. There is also a separate mode $\lambda=0$, $\psi=e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le\lambda<b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda\sinh(\eta L)=0$, leaving only that zero mode. At $\lambda=b^2$, the linear solution $\psi=A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete <Sturm-Liouville eigenfunction expansion> therefore uses
$$
\phi_0(x)=1,\quad\lambda_0=0,\qquad
\phi_n(x)=e^{-bx}\left[\cos(k_nx)+\frac b{k_n}\sin(k_nx)\right],\quad
\lambda_n=k_n^2+b^2,\quad n\ge1.
$$
Their squared <norms> in the weighted measure are
$$
N_0=\frac{e^{\beta L}-1}{\beta},\qquad
N_n=\frac L2\left(1+\frac{b^2}{k_n^2}\right),\quad n\ge1.
$$
The cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. Define
$$
K_\beta(x,y,t)=\sum_{n=0}^\infty
\frac{e^{-\lambda_nt}\phi_n(x)\phi_n(y)}{N_n}.
$$
For $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted <Lebesgue measure>. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.
To determine the boundary forcing, set $a_n(t)=\langle\phi_n,q(\cdot,t)\rangle_w/N_n$. Two applications of <integration by parts> leave
$$
N_na_n'(t)=-\lambda_nN_na_n(t)
+e^{\beta L}\phi_n(L)h_1(t)-\phi_n(0)g_1(t).
$$
The minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot]_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this <linear differential equation> by an <integrating factor> and recombining the modes gives the requested space-time <integral representation>:
$$
\boxed{q(x,t)=\int_0^L K_\beta(x,y,t)e^{\beta y}q_0(y)\,dy
+\int_0^t\left[e^{\beta L}K_\beta(x,L,t-s)h_1(s)
-K_\beta(x,0,t-s)g_1(s)\right]ds.}
$$
This is the <Neumann boundary-forcing formula>. The zero mode also gives the useful consistency check
$$
\boxed{\frac d{dt}\int_0^L e^{\beta x}q(x,t)\,dx
=e^{\beta L}h_1(t)-g_1(t).}
$$
Ordinary unweighted mass instead obeys $d\int_0^Lq\,dx/dt=h_1-g_1+\beta[q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.
There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow0$, and the prescribed derivatives are interior limits of the full solution. An explicit <Laplace transform> form verifies those limits without this interchange and gives an alternative purely contour-integral version.
For $\operatorname{Re}p>0$, set $\kappa=\sqrt{p+b^2}$ using the <principal square root>, and define
$$
u_p(x)=e^{-bx}\left[\cosh(\kappa x)+\frac b\kappa\sinh(\kappa x)\right],
$$
$$
v_p(x)=e^{-b(x-L)}\left[\cosh(\kappa(L-x))-\frac b\kappa\sinh(\kappa(L-x))\right].
$$
Both solve $(p-\mathcal A)f=0$, with $u_p(0)=v_p(L)=1$, $u_p'(0)=v_p'(L)=0$. Their weighted <Wronskian> is the nonzero constant
$$
D_p=e^{\beta x}(u_p'v_p-u_pv_p')
=e^{bL}\frac p\kappa\sinh(\kappa L).
$$
Hence the <resolvent kernel for Neumann advection-diffusion on an interval> is
$$
\mathcal R_p(x,y)=\frac{u_p(\min(x,y))v_p(\max(x,y))}{D_p}.
$$
The first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-\mathcal A_x)\mathcal R_p=e^{-\beta y}\delta_y$ in the sense of <distributional derivatives>. If $G(p)=\int_0^T e^{-ps}g_1(s)ds$ and $H(p)=\int_0^T e^{-ps}h_1(s)ds$, an equivalent transformed solution is
$$
Q(x,p)=\int_0^L\mathcal R_p(x,y)e^{\beta y}q_0(y)\,dy
+\frac{e^{\beta L}u_p(x)H(p)-v_p(x)G(p)}{D_p}.
$$
In fact $v_p'(0)=-D_p$ and $u_p'(L)=e^{-\beta L}D_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse <Bromwich contour> integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-\lambda_n$ reproduce the constant and decaying modes of $K_\beta$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.
As in question 1, the printed dot on $q_0$ denotes its spatial <derivative>: the two corner conditions are $q_0'(0)=g_1(0)$ and $q_0'(L)=h_1(0)$. The initial <eigenfunction expansion> converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac12d\|q\|_w^2/dt=-\int_0^Lw|q_x|^2dx$, proving uniqueness in the usual regular solution class. In the limit $\beta\to0$, $N_0\to L$, $\phi_n\to\cos(n\pi x/L)$, and the ordinary <Neumann heat kernel on an interval> and its boundary signs are recovered.
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