To get something beyond what you want-- keeping in mind that such multiple-subsections may need to be clubbed later on-- let's take a more general macro for collating together as many subsections as one may choose... using the \foreach loop defined in the pgffor package, and nicely elucidated in this post.
\documentclass{scrartcl}
\usepackage{pgffor, amsmath}
\renewcommand{\thesubsection}{\thesection.\alph{subsection}}
\newcommand{\multisubsection}[2]{
\subsection*{%
\thesection.%
\foreach \c in {1, ..., #1}%
{\addtocounter{subsection}{1}%
\alph{subsection},}!.
#2}}
\begin{document}
\section{The first section}
We have some questions answered in the following subsections.
\multisubsection{3}{A long answer, this should be}
When answering this question, we must consider the fact that
a complicated question expects a long and complicated, but clear answer.
\subsection{Answer to the second problem}
An easy answer, let's say.
\multisubsection{2}{Another, not-so-difficult problem}
Should have a moderately long answer, but as much to-the-point as possible.
\subsection{The next answer}
This should be evident from the above.
\end{document}
This should produce an output looking like this--

\subsection{\! b c}\addcounter{subsection}{2}. – Oni May 14 '21 at 19:39\begingroup \renewcommand{\thesubsection}{\thesection.abc} \subsection{Subsection} \endgroup \addtocounter{subsection}{2}– Ivan May 14 '21 at 20:01