Since we can't bring Suitplay or any other aid to the table (and you should not be using such things when playing online either), let's talk about how to work these out ATT
AJ9xx
xxxx Play for 4 tricks.
There are 4 cards out, which means there are 2^4= 16 possible layouts to deal with:
KQTx:- 1 x -> AJ9, cover, repeat theme
-:KQTx 1 (this one we can't do anything about, so we ignore it)
HTx:H 2 play A and then play x -> J9 OR x -> AJ9, cover, repeat theme
H:HTx 2 (another one we can't do anything about)
HT:Hx 2 x -> AJ9, play J, play A
Hx:HT 2 x -> AJ9, play J or 9, play A
KQ:Tx 1 x -> AJ9, cover, play anything to crash H+T
Tx:KQ 1 x -> AJ9, cover, play A
What becomes clear is that the line of play that covers the most layouts is
x -> AJ9, cover, repeat theme
T8x
AQ9xx Play for max tricks, only 1 entry in dummy.
32 possible layouts:
KJxxx:- 1
-:KJxxx 1
KJxx:x 3
x:KJxx 3
Kxxx:J 1
J:Kxxx 1
Jxxx:K 1
K:Jxxx 1
KJx:xx 3
xx:KJx 3
Kxx:Jx 3
Jx:Kxx 3
Jxx:Kx 3
Kx:Jxx 3
KJ:xxx 1
xxx:KJ 1
Using the method shown for the 1st problem, find the line of play that leads to Max tricks for as many of these 32 layouts as possible assuming you can only get to T8x once to take any finesses.
Answer posted later if necessary.