If I understand correctly, the problem is to reduce
$$
(\overline{A \& B} \& C) + (\overline{A} \& B \& C)+ (\overline{A \& B} \& C) +(A\& B \& C)
$$
You can deduce from the laws that $X+X=X$, so this is clearly already
$$
=(\overline{A \& B} \& C) + (\overline{A} \& B \& C)+(A\& B \& C)
$$
By law 3, then law 1, then law 2: $ (\overline{A} \& B \& C)+(A\& B \& C)=(\overline{A}+A)\&B\& C=1\&B\& C=B\&C $, so the original expression is now:
$$
=(\overline{A \& B} \& C)+B\&C
$$
By De Morgan's laws $\overline{A \& B}=(\overline{A}+\overline{B})$, and you can deduce the rest from your laws:
$$
=(\overline{A} + \overline{B}) \& C+B \& C =(\overline{A} + \overline{B}+B) \& C
$$
$$
=(\overline{A} + 1) \& C=1\&C=C
$$
HINT
This equivalence principle will be your friend:
Adjacency
$PQ + PQ' = P$
If you're not allowed to use Adjacency in $1$ step, here is a derivation of Adjacency in terms of more basic equivalence principles:
$$P Q + (P Q') \overset{Distribution}=$$
$$P (Q + Q') \overset{Complement}=$$
$$P 1 \overset{Identity}=$$
$$P$$
To apply Adjacency, note that $P$ and $Q$ can be any complex expressions, so in this case, where every terms has $4$ variables, just look for two terms that are the same for $3$ of the variables, but differ in the fourth. For example, the first two terms are the same except for the $D$ variable, so those can be combined:
$A'BC'D'+A'BC'D=A'BC'$
You can also combine the first and seventh terms:
$A'BC'D'+ABC'D'=BC'D'$
To do both of those, you would need to 'reuse' the first term, but you can get as many copies as you want by:
Idempotence
$P + P = P$
So, for example, focusing on the first, second, and seventh term:
$$A'BC'D'+A'BC'D+ABC'D'\overset{Idempotence}=$$
$$A'BC'D'+A'BC'D'+A'BC'D+ABC'D'\overset{Commutation}=$$
$$A'BC'D'+A'BC'D+A'BC'D'+ABC'D'\overset{Adjacency \ x \ 2}=$$
$$A'BC'+BC'D'$$
Best Answer
Karnaugh maps are your best friend. Otherwise, observe that:
$$ \begin{align*} H &= M'CD' + M + CRD \\ &= M'CD' + M(1) + CRD \\ &= M'CD' + M(CD'+1) + CRD \\ &= M'CD' + MCD' + M + CRD \\ &= (M' + M)CD' + M + CRD \\ &= (1)CD' + M + CRD \\ &= CD' + M + CRD \\ &= CD' + CRD + M \\ &= C(1)D' + CRD + M\\ &= C(1 + R)D' + CRD + M\\ &= CD' + CRD' + CRD + M\\ &= CD' + CR(D' + D) + M\\ &= CD' + CR(1) + M\\ &= CD' + CR + M\\ \end{align*} $$