I believe $T(5,4)=55$.
We have a convex pentagon in the plane. Choose a point one each side; call those points $1,2,3,4,5$ in order around the pentagon. By previous discussion, we want to count all the non-self-intersecting spanning trees one can draw on these 5 points, using only line segments between pairs of the points.
There are, up to isomorphism, 3 trees on 5 points; I'll call them X, Y, and I (all to be imagined sans-serif).
There are 5 ways to implement the X; choose any one of the 5 points to be the vertex of order 4, then you have only one way to draw the tree.
To implement the Y, first pick a point to be the vertex of order 3; 5 ways. For concreteness, let's say we have chosen point 1. Now choose the three points to connect to 1. There are 4 ways to do this. If you choose $2,3,4$, then you must connect 4 to 5; if you choose $3,4,5$, then you must connect 2 to 3; that's 10 trees. If you choose $2,3,5$, then you can connect 4 to either 3 or 5; if you choose $2,4,5$, then you can connect 3 to either 2 or 4; that's another 20 trees. All told, 30 ways to implement the Y.
To implement the I, note that the endpoints will either be on adjacent edges of the pentagon, or on edges separated by one edge; we'll call these cases A and B. In either case, we can distinguish one endpoint as the start, the other as the end, by agreeing that the clockwise distance around the pentagon from start to end must be less than half way round the pentagon.
Case A: 5 ways to choose the start, and after that only one way to complete the tree (e.g., if the start is 1, so the end is 2, you must connect 1 to 5, 5 to 4, 4 to 3, and 3 to 2).
Case B: 5 ways to choose the start, and 3 ways to complete the tree. E.g., if the start is 1, so the end is 3, then you can go 1-2-5-4-3, 1-5-2-4-3, or 1-5-4-2-3. (This whole exercise would be so much easier if I knew how to draw and post diagrams).
So, 5 trees in Case A, 15 in Case B, making 20 ways to implement the I, making $T(5,4)=5+30+20=55$
EDIT: Now we have enough data to go to the Online Encyclopedia of Integer Sequences, and we find http://oeis.org/A001764 which is said to enumerate non-crossing trees and to be given by ${1\over2n+1}{3n\choose n}$
EDIT 29 July 11: I reckon $T(5,3)=15$.
We have 5 arcs, and 6 ends-of-chords, so one arc has two ends-of chords, the other arcs have one end-of-chord each.
Label the arcs $1,2,3,4,5$ as for the $T(5,4)$ calculation. Assume the arc with two ends-of-chords is arc 1 (then multiply whatever number we get by 5).
The two chords ending on arc 1 could go to arcs 3 and 4. Then the third chord must join arcs 2 and 5.
The two chords ending on arc 1 could go to arcs 2 and 4. Then the third chord must join arcs 3 and 5.
The two chords ending on arc 1 could go to arcs 3 and 5. Then the third chord must join arcs 2 and 4.
No other placement of the two chords ending on arc 1 permits a third chord that separates the vertices, so there are just these three ways. $3\times5=15$.