How can you really be sure the contradiction didn't spring from the hypothesis? This question may have a duplicate but I didn't find one.
Given a proof by contradiction of a statement like $p \land q \implies r$. Which means (as i understand it):
$p \land q \land \lnot r$ is false.
(1) Is it right that for the proof to be valid one should verify p∧q is true?
ADDED: (2) Is it right that for the proof to be sound one should verify $p \land q$ doesn’t lead to a contradiction?
 A: No, not necessarily. For a proof to be sound, then indeed, we need for $p\land q$ to be true.
But validity depends only on the form of an argument, not the semantic content of the argument. An argument or proof is valid provided the conclusion is logically implied by the premises. Indeed, we can have valid argument forms in which one or more premises is false.

If my apple is blue, then pigs can fly.
My apple is blue.
Therefore, pigs can fly.

This is a valid argument but it certainly is not sound: It is valid by modus ponens. All arguments of the form $$\begin{align} P \rightarrow Q \\ \\ P \\ \hline \\ \therefore Q\end{align}$$
are valid, just by virtue of the structure or form of the argument.
Of course, mathematical proofs usually only become interesting when they are sound: when the premises are true, and the conclusion(s) logically follow(s) from the premises. So indeed, we take great care to ensure that the premises are true, which would ensure us that the premises of an argument, by themselves, are not responsible for any contradiction reached when assuming the negation of the desired conclusion!
A: It is wrong. What you are describing in your last line, is the modus ponens, where from $A \to B$ and $A$ one deduces $B$. But the truth of $A$ has nothing to do with the truth of the implication $A \to B$, in fact this is completely true:

If $1$ is $0$, then $1$ is $1$

