= Solution
Choose a <base of a root system> and its <positive roots>, so that the <triangular decomposition of a Lie algebra> is $L=\mathfrak n^-\oplus H\oplus\mathfrak n^+$. A <primitive element of a Lie algebra representation> of weight $\omega\in H^*$ is a nonzero vector $v$ such that
$$
hv=\omega(h)v\quad(h\in H),\qquad xv=0\quad(x\in\mathfrak n^+).
$$
Equivalently it is a <highest-weight vector>. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.
Construct the <Verma module> for an arbitrary $\omega$. The subalgebra $\mathfrak b=H\oplus\mathfrak n^+$ has a one-dimensional <module> $\mathbb C_\omega$ on which $H$ acts by $\omega$ and $\mathfrak n^+$ acts by zero. This is a representation because $[\mathfrak b,\mathfrak b]\subseteq\mathfrak n^+$. Set
$$
M(\omega)=U(L)\otimes_{U(\mathfrak b)}\mathbb C_\omega,
$$
where $U$ denotes the <universal enveloping algebra>. The <Poincaré-Birkhoff-Witt theorem>, applied with negative-<root vectors> first, then Cartan vectors and positive-<root vectors>, gives
$$
M(\omega)\cong U(\mathfrak n^-)
$$
as vector spaces. In particular $v_\omega=1\otimes1$ is nonzero, generates $M(\omega)$, and is primitive of weight $\omega$.
The same ordered monomials show that $M(\omega)$ is a <direct sum> of weight spaces, with weights of the form
$$
\omega-\sum_{i=1}^r n_i\alpha_i,\qquad n_i\in\mathbb Z_{\ge0},
$$
and that its weight-$\omega$ space is exactly $\mathbb Cv_\omega$. Each vector has only finitely many weight components. If a <submodule> contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. <Polynomial interpolation> in the action of $h$ extracts a nonzero multiple of $v_\omega$. Such a <submodule> is all of $M(\omega)$, because $v_\omega$ generates it. Consequently every proper <submodule> has zero component in weight $\omega$.
Let $N$ be the sum of all proper <submodules>. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper <submodule> and is therefore the unique maximal proper <submodule>. Its quotient
$$
\boxed{L(\omega)=M(\omega)/N}
$$
is irreducible, and the nonzero image of $v_\omega$ is primitive of weight $\omega$. This proves existence for every $\omega\in H^*$ and gives the <irreducible quotient of a Verma module>. It also proves uniqueness up to isomorphism among irreducible <modules> generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega)$, whose proper kernel must be $N$.
These <modules> are not generally finite-dimensional. For example, in an $\mathfrak{sl}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfies
$$
e f^k v=k(\lambda-k+1)f^{k-1}v.
$$
If its irreducible highest-weight <module> is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^mv\ne0$, the identity forces $\lambda=m\in\mathbb Z_{\ge0}$. In general a finite-dimensional highest-weight <module> therefore requires nonnegative integral values on all <simple coroots>; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.
For <simple coroots> $h_1,\ldots,h_r$ forming a basis of $H$, the <fundamental weights> are the <dual basis> $\omega_1,\ldots,\omega_r\in H^*$:
$$
\omega_i(h_j)=\delta_{ij}.
$$
For the <special linear Lie algebra> $\mathfrak{sl}_3(\mathbb C)$, take the diagonal <trace>-zero <Cartan subalgebra> and <positive roots> corresponding to upper triangular matrix units. Write $\varepsilon_i(\operatorname{diag}(t_1,t_2,t_3))=t_i$. The <simple roots> are $\alpha_1=\varepsilon_1-\varepsilon_2$, $\alpha_2=\varepsilon_2-\varepsilon_3$, and the <simple coroots> are
$$
h_1=\operatorname{diag}(1,-1,0),\qquad h_2=\operatorname{diag}(0,1,-1).
$$
The <fundamental weights of sl3> are therefore
$$
\boxed{\omega_1(\operatorname{diag}(t_1,t_2,t_3))=t_1,\qquad
\omega_2(\operatorname{diag}(t_1,t_2,t_3))=t_1+t_2=-t_3}.
$$
Checking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean <trace>-zero plane they can also be written
$$
\omega_1=\frac{2\alpha_1+\alpha_2}{3}=\left(\frac23,-\frac13,-\frac13\right),\qquad
\omega_2=\frac{\alpha_1+2\alpha_2}{3}=\left(\frac13,\frac13,-\frac23\right).
$$
The <Fundamental representations of sl3> make these weights concrete: $\mathbb C^3$ has primitive vector $e_1$ of weight $\omega_1$, and $\Lambda^2\mathbb C^3$ has primitive vector $e_1\wedge e_2$ of weight $\omega_2$. Every finite-dimensional irreducible highest weight is $a\omega_1+b\omega_2$ with nonnegative integers $a,b$.
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