Those took me a while.

A.B + B.!C + !A.!C

A.B + !A.!C + B.!C // reordered -> the last one is some sort of linear combination of the other two
A.B + !A.!C + (B.!C).(A+!A) // augmented with something true (A+!A)
A.B + !A.!C + B.!C.A + B.!C.!A // expanded
A.B + B.!C.A + !A.!C + B.!C.!A // reordered
(A.B).(1+!C) + (!A.!C).(1+B) // factored out our "basis" (we knew that the third term in the beginning was a combination of those two)
(A.B) + (!A.!B) // done.

D + !B.!C + C.!D + !A.!B.C

(D + C.!D) + (!B.!C + !A.!B.C) // reordered. I placed brackets around the two groups we simplify next.
(D+C).(D+!D) + (!B.(!C + (!A.C))) // factor out
(D+C) + (!B.(!C+!A).(!C+C)) // left: identity, right: expanded
(D+C) + (!B.(!C+!A)) // right: identity removed,
D+C + !B.!C + !B.!A // expanded
D+(C+ !B.!C) + !B.!A // again I placed brackets to show what will be simplified next
D+(C+!B).(C+!C)) +!B.!A // once more we found an identity
D+C+ (!B+ !B.A) // and finally the last simplification brings us to:
D+C+!B // done.

Cheers!

PS: You can verify the steps with wolframalpha.com!

The way I approach these problems, I try to visualize the truth table of the expression and then identify large chunks that are 1. For instance,

A.B + B.!C + !A.!C


C=0:
B=0 B=1
A=0 1 1
A=1 0 1

C=1:
B=0 B=1
A=0 1 0
A=1 0 0


Try to visualize one table in front of the other, forming a cube. You can now group the ones in two groups: one is A.B and the other is B.!C. So the expression is simply A.B + B.!C

Sometimes it helps to transfer the truth table to a Karnaugh map. (Makes it easier to see groups.)