It’s not too hard to check the results after you know them:
$$\begin{align*}B'D+ABC'D&=(B'+ABC')D\\
&=(B'+AB'C'+ABC')D\\
&=\Big(B'+A(B'+B)C')\Big)D\\
&=(B'+AC')D\\
&=B'D+AC'D
\end{align*}$$
and
$$\begin{align*}A'BC + B'C + AC&=B'C+(A'B+A)C\\
&=B'C+(A'B+A+AB)C\\
&=B'C+\Big(A+(A'+A)B\Big)C\\
&=B'C+(A+B)C\\
&=B'C+AC+BC\;.
\end{align*}$$
That second one can be further simplified to $AC+C=C$, since $B'C+BC=(B'+B)C=C$.
In both calculations I used the absorption law: $B'=B'+AB'C'$ in the first, and $A=A+AB$ in the second.
When you have no more than three or four proposition letters, you may find Venn diagrams helpful.
You can not use associativity when operators are mixed:
$$a'.b'.c' + a.b'.c' + a.b.c' \neq (b'.c').(a'+ a) + a.b.c'\tag{1}$$
It is true that $a'.b'.c' = b'.c'.a'$, by commutativity of ".", but it is not legitimate/not valid to impose parentheses to group $(a' + a)$ as you did.
You can use the distributive laws, and you might want to use Demorgan's Laws, as well.
Example: for using the distributive law, and the fact that $b' + b = 1$
$$a.b'.c' + a.b.c' = a.c'.b' + a.c'.b = a.c'(b'+b) = a.c'$$
Now we've simplified our expression to
$$a'.b'.c' + a.c'\tag{2}$$
There's a common term of $c'$ in each of these products, so we can simplify further. See what you come up with, and I'll work with you to clarify/confirm, etc. if you have any more questions.
$$a'.b'.c' + a.c'=(a'.b' + a).c' $$ $$= [(a'+a).(b'+a)].c' $$ $$= (b'+a).c' = b'.c' + a.c' $$ $$= a.c' + b'.c'$$
Best Answer
It always helps to expand the terms so they include all variables .. and then reorganize, and recombine in a more efficient way, adding or removing duplicates as needed.
The key principle is:
Adjacency
$P = PQ + PQ'$
Starting from your second expression:
$(AB+A'D+B'D)C=$
$(ABD+ABD'+A'BD+A'B'D+AB'D+A'B'D)C=$
$(ABD+ABD'+(AB+A'B+AB'+A'B')D)C=$
$(AB+(B+B')D)C=$
$(AB+D)C$