= Solution
The <irregularity of an algebraic surface> and the <geometric genus of an algebraic surface> are respectively
$$
q(X)=h^1(X,\mathcal O_X)=h^0(X,\Omega_X^1),
\qquad
p_g(X)=h^0(X,K_X)=h^2(X,\mathcal O_X).
$$
If $\pi:X'\to X$ is the blowup at a point with <exceptional divisor> $E$, then
$$
K_{X'}=\pi^*K_X+E.
$$
A holomorphic two-form on $X$ pulls back to one on $X'$. Conversely, a holomorphic two-form on $X'$ descends across $E$: locally it is a form on the punctured smooth surface $X\setminus\{p\}$, and its coefficients extend over the codimension-two point $p$. Equivalently, $\pi_*K_{X'}=K_X$. Pullback is therefore an isomorphism
$$
H^0(X,K_X)\cong H^0(X',K_{X'}),
$$
which proves the <birational invariance of the geometric genus of a surface> in this case.
Let $X=C\times D$, where both smooth projective curves have positive genus. The <Künneth theorem> gives the <irregularity of a product of curves>
$$
q(X)=g(C)+g(D)>0.
$$
If a surface is a hypersurface in projective space, dimension forces it to be a smooth hypersurface $Y\subseteq\mathbb P^3$. From
$$
0\longrightarrow\mathcal O_{\mathbb P^3}(-d)
\longrightarrow\mathcal O_{\mathbb P^3}
\longrightarrow\mathcal O_Y
\longrightarrow0
$$
and the intermediate cohomology vanishing for line bundles on projective space, one obtains $H^1(Y,\mathcal O_Y)=0$. Thus every such hypersurface has irregularity zero, whereas $X$ has positive irregularity. Hence $C\times D$ is not isomorphic to a hypersurface. This is the <product of positive-genus curves is not a projective hypersurface> obstruction.
The <Albanese variety> $\operatorname{Alb}(X)$ is the universal <abelian variety> receiving a pointed morphism from $X$, and
$$
\dim\operatorname{Alb}(X)=q(X).
$$
Let the smooth image curve of the Albanese morphism be $C$ of genus $g$. Pullback of holomorphic one-forms along the dominant map $X\to C$ is injective, giving $g\leq q(X)$. On the other hand, the universal property of the Jacobian extends $C\to\operatorname{Alb}(X)$ to a homomorphism
$$
\operatorname{Jac}(C)\longrightarrow\operatorname{Alb}(X).
$$
The Albanese image generates the entire Albanese variety, so this homomorphism is surjective and $q(X)\leq\dim\operatorname{Jac}(C)=g$. Therefore the <Irregularity from a smooth Albanese curve image> is
$$
\boxed{q(X)=g}.
$$
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