Let us construct the truth tables for the statements:
$(1)\; (p \wedge q)\rightarrow (p \vee q)$
$\begin{matrix} p & q & p \wedge q & p \vee q & p \wedge q \rightarrow p \vee q \\ T & T & T& T & T \\ T & F & F& T & T \\ F & T & F & F & T\\ F & F & F & F & T \end{matrix}$
Therefore, this is a tautology
$ (2)\; (p\wedge(\sim p))\wedge((\sim q)\wedge p)$
$\begin{matrix} p & q & \sim p & \sim q & p \vee (\sim q) & (\sim q)\vee p & (p\wedge(\sim p))\wedge((\sim q)\wedge p) \\ T & T & F& F & F & F & F\\ T & F & F & T & F & T & F\\ F& T & T & F & F & F & F\\ F& F & T & T& F & F & F \end{matrix}$
Therefore, this is a contradiction.
$(3) \; q\vee (p\vee(\sim q)) $
$\begin{matrix} p & q & \sim q & p \vee (\sim q) &q\vee (p\vee(\sim q)) \\ T&T & F& T &T \\ T & F & T &T &T \\ F& T & F & F &T \\ F& F & T & T & T \end{matrix}$
Therefore, this is a tautology.
Therefore the correct answer is (C) Only Statement (2) is a contradiction, (1) and (3) are tautologies.