= Solution
A <minimal algebraic surface> is a smooth projective surface containing no <exceptional curve of the first kind>, namely no smooth rational curve of self-intersection $-1$. An <abelian surface> contains no rational curve because every morphism from $\mathbb P^1$ to an <abelian variety> is constant. It therefore has no $(-1)$-curve and is minimal.
For every $n\geq2$, the <Minimal Hirzebruch surface> $\mathbb F_n$ is a rational minimal surface. These surfaces are pairwise nonisomorphic: the negative section is the unique irreducible curve of negative self-intersection and has square $-n$, so an isomorphism would recover $n$. Thus there are infinitely many nonisomorphic minimal rational surfaces.
A <K3 surface> is a smooth projective surface $X$ with $K_X\cong\mathcal O_X$ and $H^1(X,\mathcal O_X)=0$. For a smooth curve $C\subset X$ of <geometric genus> $g$, the <Adjunction formula> gives
$$
2g-2=C\cdot(C+K_X)=C^2,
$$
and hence
$$
\boxed{C^2=2g-2}.
$$
An <elliptic surface> is a smooth projective surface with a morphism to a smooth curve whose generic fiber is a smooth genus-one curve. If $E$ is an <elliptic curve>, projection
$$
\mathbb P^1\times E\longrightarrow\mathbb P^1
$$
is an elliptic fibration. Its canonical bundle is pulled back from $K_{\mathbb P^1}$, so every positive pluricanonical space vanishes and the <Kodaira dimension> is $-\infty$. This supplies the requested negative-Kodaira-dimension example.
For an elliptically fibered K3 surface, choose a smooth quartic $X\subseteq\mathbb P^3$ containing a line $L$. The <canonical bundle of a smooth projective hypersurface> formula makes $K_X$ trivial, and the standard cohomology sequence gives $H^1(X,\mathcal O_X)=0$, so $X$ is K3. The pencil of planes through $L$ cuts $X$ into $L$ plus a residual plane cubic. The residual linear system $|H-L|$ is basepoint-free, has square zero and defines a morphism $X\to\mathbb P^1$ whose generic fiber is a smooth plane cubic. This is the <Elliptic K3 surface from a quartic containing a line>.
A <surface of general type> is a smooth projective surface of <Kodaira dimension> two. Let $B\subseteq\mathbb P^2$ be a smooth plane curve of degree eight and let
$$
\pi:X\longrightarrow\mathbb P^2
$$
be the degree-two cover branched along $B$. The branch-cover canonical-bundle formula gives
$$
K_X=\pi^*\left(K_{\mathbb P^2}+4H\right)=\pi^*H.
$$
This divisor is ample, so $X$ is a <double plane of general type> and $\pi$ is the required finite morphism of degree two.
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