Boolean expression simplification using 3 variables (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABD)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABD)(1)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABD)(C + !C)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABDC) + (!ABD !C)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABDC) + (!ABD !C)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABCD) + (!AB !CD)
= (!A B !C) + (!AB !CD) + (B !C !D) + (!ACD) + (!ACDB) + (!BCD)
= [(!A B !C)(1 + D)] + (B !C !D) + [(!ACD)(1 + B)] + (!BCD)
= (!A B !C) + (B !C !D) + (!ACD) + (!BCD)
At this point I don't know how to continue the problem.
I verified it using a calculator and seems there is still some simplification to do.
 A: Here are two handy equivalences:
Adjacency
$PQ + P!Q=P$
Absorption
$P + PQ=P$
With that:
$$(!A B !C) + (B !C !D) + (!ACD) + (!BCD) + (!ABD)\overset{Adjacency \ x \ 2}=$$
$$(!A B !C D) + (!A B !C !D) + (B !C !D) + (!ABCD) + (!A!BCD)+ (!BCD) + (!ABD)\overset{Absorption \ x \ 4}=$$
$$(B !C !D) + (!BCD) + (!ABD)$$
I am sure that went a little quick though, so here is an explanation:
In step 1, I applied Adjacency to:
$(!A B !C)$ ... becoming $(!A B !C D) + (!A B !C !D)$ 
and to $(!ACD)$ ... becoming $(!ABCD) + (!A!BCD)$
Then, in step 2:
$(!A B !C D)$ got absorbed by $(!ABD)$
$(!A B !C !D)$ got absorbed by $(B !C !D)$
$(!ABCD)$ got absorbed by $(!ABD)$
$(!A!BCD)$ got absorbed by $(!BCD)$
More equivalences to put in your toolbox! 
Now, notice that you effectively did Adjacency as well in your first few steps, expanding $(!ABD)$ to $(!ABCD)+(!AB!CD)$ ... and yet that one isn't as useful as the ones I did. So, I know you are going to ask: how did I know which terms to expand using Adjacency? Well, I created a little K-map. Look it up!
