Whist schedules only cover 4n and 4n+1 players, and so it gets difficult when the numbers of players is 4n+2 or 4n+3. To see the problem, count the total number of doubles partnerships that it is possible to have. For example with 4n+2 players there are 8n^2 + 6n + 1 possible partnerships, which will always be an odd number; however each doubles game involves exactly 2 partnerships, and therefore any schedule that you care to make will have an even number of partnerships. Therefore you can never have a schedule where everybody plays with everyone else exactly once.
To make any progress you need to decide where to compromise. The smallest change would be to assign one pair of players who never play as partners or opponents, but a consequence is that this pair will have one game fewer than all other players. For example consider the following 22 game schedule:
10 players A to J, where the pair (A F) never play in the same game together.
[(A E):(B H)] [(D F):(C I)]
[(B I):(D H)] [(C G):(E J)]
[(B D):(A C)] [(H J):(F I)]
[(A G):(H I)] [(D E):(B F)]
[(I J):(B E)] [(G H):(C D)]
[(B C):(G I)] [(D J):(E H)]
[(E F):(D G)] [(A B):(C J)]
[(G J):(A D)] [(C F):(E I)]
[(A H):(C E)] [(F G):(B J)]
[(E G):(A I)] [(F J):(C H)]
[(A J):(D I)] [(B G):(F H)]
For the players other than A & F, the experience will be that of a perfect Whist tournament, all partners once, all opponents twice.
In general when one pair has a game less, the cyclic methods using starter blocks are no longer going to be of any use, as they always lead to schedules where everyone has the same number of games. In fact the schedule above was found with a fairly crude computer search that attempts to arrange the 44 partnerships into a schedule. I have not found any similar schedule for more than 10 players, that can be arranged into rounds of simultaneous play, like the one above.
If it is essential to give everyone the same number of games, the only option I see is to arrange it so that everyone has some other player with whom they never meet. Would that be an option? So for example with 10 players there would be only 20 games with 40 out of the possible 45 partnerships.
Taking your three points in turn.
(1) I mean both at the same time. That there is one person with whom you never partner and never oppose. This idea is similar to"Spouse Avoidance" which is something usually associated with mixed doubles. Here's an example for 10 players:
(A B v E I) (D J v F G)
(B C v A J) (E F v G H)
(C D v B F) (A G v H I)
(D E v C G) (B H v I J)
(E A v D H) (C I v J F)
(E G v F I) (A D v B J)
(A H v G J) (B E v C F)
(B I v H F) (C A v D G)
(C J v I G) (D B v E H)
(D F v J H) (E C v A I)
where the pairs who never meet are AF, BG, CH, DI & EJ. Send me an email by clicking my name if you are interested in more like this.
(2) For 4n-1 you cannot just delete one player, unless you are prepared to play two against one. Which for tennis is presumably something you would not want to do. The alternative is to remove the whole game, not just the deleted player, but then you will be removing one partnership and two opposition pairs, that are not associated with the deleted player.
(3) Whist for 4n+1 is not covered on this site, but they are well known. For example see the "pure" individual pairs schedules here (http://www.jdawiseman.com/papers/tournaments/individual-pairs/individual-pairs.html). The asymmetric schedules may also be of interest.