Answer
$(\neg p\wedge(p\rightarrow q))\rightarrow\neg q$ is not a tautology
Work Step by Step
$\left[\begin{array}{ll}
p & q\\
\hline & \\
T & T\\
T & F\\
F & T\\
F & F
\end{array}\right. \left|\begin{array}{lll}
\neg p & p\rightarrow q & \neg p\wedge(p\rightarrow q)\\
\hline & & \\
F & T & F\\
F & F & F\\
T & T & T\\
T & T & T
\end{array}\right|\left.\begin{array}{ccc}
\neg q & (\neg p\wedge(p\rightarrow q))\rightarrow\neg q\\
\hline & \\
F & T\\
T & T\\
F & F\\
T & T
\end{array}\right]$
$(\neg p\wedge(p\rightarrow q))\rightarrow\neg q $ does not have all T values in its column,
so it is not a tautology.
(in row 3, $T\rightarrow F$ is false)
