I'm studying for my upcoming discrete math test and I'm having trouble understanding some equivalences...

Step 1
The only step that should cause any trouble is the first, the equivalence of
$(\mathrm{\neg }P\wedge \mathrm{\neg }Q)\vee (\mathrm{\neg }Q\wedge \mathrm{\neg }R)$
with $(\mathrm{\neg }P\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R))\wedge (\mathrm{\neg }Q\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R));$
from that to $(\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)$
is just removing redundant parentheses, and from that to $(\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }R)$
just uses the fact that $X\vee X$ is equivalent to X.
The first step is actually impossible as it stands: if Q is false and P and R are true, the second line is true, but the first is false. It appears from the bulk of the problem that the first line was supposed to read $(\mathrm{\neg }P\wedge \mathrm{\neg }Q)\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R);$
if that is indeed the case, the first step is just an application of distributivity of ∨ over ∧, i.e., of the equivalence of $(X\wedge Y)\vee Z$ with $(X\vee Z)\wedge (Y\vee Z)$; $\mathrm{\neg }P$ is the X, $\mathrm{\neg }Q$ is the Y, and $(\mathrm{\neg }Q\vee \mathrm{\neg }R)$ is the Z.

Step 1
(1) Put the argument in adequate symbolism:
Statements in ordinary language like the above reasoning may be misleading, ambiguous confusing. Therefore, let us prove it an adequate symbollism. Let '¬' stand for negation, '∧' for conjunction and '∨' for disjunction. Is the reasoning below what you mean?
$\begin{array}{r}\text{(1)}& (\mathrm{\neg }P\wedge \mathrm{\neg }Q)\vee (\mathrm{\neg }Q\wedge \mathrm{\neg }R)& \equiv (\mathrm{\neg }P\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R))\wedge (\mathrm{\neg }Q\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R))\text{(2)}& & \equiv (\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\text{(3)}& & \equiv (\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }R)\end{array}$
One might say that the first step (1) is an application of distributivity of ∨ over ∧. I am afraid this would be incorrect.
As we know, the distributivity of ∨ over ∧, in symbols:
$\models (\varphi \wedge \psi )\vee \sigma \equiv (\varphi \vee \sigma )\wedge (\psi \vee \sigma )$
should state that:
$\begin{array}{}\text{(1*)}& (\underset{\varphi }{\underbrace{\mathrm{\neg }P}}\wedge \underset{\psi }{\underbrace{\mathrm{\neg }Q}})\vee \underset{\sigma }{\underbrace{(\mathrm{\neg }Q\wedge \mathrm{\neg }R)}}& \equiv (\underset{\varphi }{\underbrace{\mathrm{\neg }P}}\vee \underset{\sigma }{\underbrace{(\mathrm{\neg }Q\wedge \mathrm{\neg }R)}})\wedge (\underset{\psi }{\underbrace{\mathrm{\neg }Q}}\vee \underset{\sigma }{\underbrace{(\mathrm{\neg }Q\wedge \mathrm{\neg }R)}})\end{array}$
Which as we see in the formalism above, is not the case. In fact (1) is not a logical equivalence at all (you can check it by a simple truth table).
(2) Typo:
We probably have a typo in the OP's question: presumably, it was supposed to be either
$(notPandnotQ)or(notQandnotR)$
$\Leftarrow \Rightarrow (notPor(notQandnotR))and(notQor(notQandnotR))$
$\Leftarrow \Rightarrow (notPornotQandnotR)and(notQornotQandnotR)$
$\Leftarrow \Rightarrow (notPornotQandnotR)and(notQandnotR)$
or $(notPandnotQ)or(notQornotR)$
$\Leftarrow \Rightarrow (notPor(notQornotR))and(notQor(notQornotR))$
$\Leftarrow \Rightarrow (notPornotQornotR)and(notQornotQornotR)$
$\Leftarrow \Rightarrow (notPornotQornotR)and(notQornotR)$
Given the argument's structure, I bet in the second one (the first one is nonsense). In this case, we have the following correction of the formal reasoning above:
$\begin{array}{r}\text{(1')}& (\mathrm{\neg }P\wedge \mathrm{\neg }Q)\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R)& \equiv (\mathrm{\neg }P\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R))\wedge (\mathrm{\neg }Q\vee (\mathrm{\neg }Q\vee \mathrm{\neg }R))\text{(2')}& & \equiv (\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\text{(3')}& & \equiv (\mathrm{\neg }P\vee \mathrm{\neg }Q\vee \mathrm{\neg }R)\wedge (\mathrm{\neg }Q\vee \mathrm{\neg }R)\end{array}$
In (1') we have an application of the distributivity of ∨ over ∧.
In (2') we are justified by the associativity of ∨.
In (3') we have an application of the idempotency of ∨.