= Solution
A <simple module> is nonzero and has no <submodules> other than zero and itself. A <semisimple module> is a <direct sum> of simple <submodules>; the zero <module> is allowed as the empty sum.
For (1)$\Rightarrow$(2), argue by <dimension>. Choose a minimal nonzero <submodule> $S$, which is simple, and split $V=S\oplus C$ using (1). The property passes to $C$: for $U\subseteq C$, write $V=U\oplus D$; then $C=U\oplus(C\cap D)$. Apply induction to $C$ and append $S$. The zero-dimensional case starts the induction. The implication (2)$\Rightarrow$(3) is immediate. For (3)$\Rightarrow$(2), repeatedly append a simple <submodule> not contained in the sum already chosen. Its intersection with that sum is zero by simplicity. <Dimension> strictly increases, so after finitely many steps their <direct sum> is $V$.
For (2)$\Rightarrow$(1), write $V=\bigoplus_iS_i$ and fix $U$. Choose a sum $C$ of a maximal subset of these $S_i$ whose sum is disjoint from $U$. If $U+C\ne V$, some $S_j$ is not contained in $U+C$, so simplicity gives $S_j\cap(U+C)=0$. Then $C+S_j$ is a larger <direct sum> still disjoint from $U$, a contradiction. Hence $V=U\oplus C$.
Condition (1) implies (1\*). Conversely suppose (1\*) and let $W$ be the sum of all simple <submodules>. If $W\ne V$, finite <dimension> supplies a maximal proper <submodule> $M$ containing $W$. By (1\*), $V=M\oplus C$. Projection to the simple quotient $V/M$ identifies the nonzero complement $C$ with a <simple module>. Thus $C\subseteq W\subseteq M$, contradicting the <direct sum>. Consequently $W=V$, which is (3). This proves \b[all four conditions are equivalent].
For the concrete complement, $x^2=x$. <Orthogonal idempotents> $x$ and $1-x$ give $A=Ax\oplus A(1-x)$: the sum follows from $1=x+(1-x)$, and right multiplication by $x$ shows their intersection is zero. In explicit <matrix> form,
$$
\boxed{U=\left\{\begin{pmatrix}a&\lambda a\\b&\lambda b\end{pmatrix}:a,b\in K\right\},\qquad C=\left\{\begin{pmatrix}0&c\\0&d\end{pmatrix}:c,d\in K\right\}=A(1-x).}
$$
The decomposition of $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ uses second-column remainders $b-\lambda a,d-\lambda c$ in $C$.
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