a.
There are $3$ triangles having the same number on each side ($111$, $222$, $333$). There are $6$ triangles having one number on two sides and another number on the third side ($112$, $113$, $221$, $223$, $331$, $332$). Only $1$ triangle has a different number on every side ($123$). Thus there are $10$ different triangles consisting of three sticks in total.

b.
The largest sum of numbers written on $9$ sticks is $27$. Suppose that the sum of the numbers on the sticks on the boundary of the big triangle is $27$. This means that all sticks on the boundary have $3$ on it. There are $6$ different triangles consisting of three sticks having $3$ on some side. As the number of small triangles having a side on the boundary of the big triangle is also $6$, all $6$ different triangles consisting of three sticks and having $3$ on some side must occur at the corners or in the middle of a side of the big triangle. As one of these $6$ triangles has $3$ on every side, but not all sides of a small triangle can lie on the boundary of the big triangle, at least one stick with number $3$ lies in the interior of the big triangle. The other small triangle with this stick as a side lies entirely in the interior of the big triangle. This contradicts the previously proved claim that all triangles consisting of three sticks and having $3$ on one side lie on the boundary of the big triangle. The contradiction shows that the sum $27$ is impossible.

Figure 24 shows that the sum can be $26$.

Any of Figures 25, 26, 27, 28 and 29 shows that the sum can be $26$.

*Remark:* Figures 24–29 contain all possibilities, modulo rotations and reflections, for obtaining the sum $26$.

a.
The question of the problem is equivalent to the question how many three-digit numbers whose each digit is $1$, $2$ or $3$ and digits are in non-decreasing order do there exist. There are $10$ such numbers ($111$, $112$, $113$, $122$, $123$, $133$, $222$, $223$, $233$, $333$).

b.
The largest sum of numbers written on $9$ sticks is $27$. Suppose that the sum of the numbers on the sticks on the boundary of the big triangle is $27$. This means that all sticks on the boundary have $3$ on it. In all $10$ triangles consisting of three sticks, the number $3$ occurs $10$ times in total. To have $9$ occurrences on the boundary, the triangle with $3$ on every side must definitely be used. As not all sides of a small triangle can lie on the boundary of the big triangle, at least one stick with number $3$ lies in the interior of the big triangle. Thus all $10$ occurrences must be used. But the stick in the interior of the big triangle and having $3$ on it is a side of another triangle entirely in the interior of the big triangle, whence we have $11$ occurrences of $3$ when counted by triangles. The contradiction shows that the sum $27$ is impossible.

Any of Figures 25, 26, 27, 28 and 29 shows that the sum can be $26$.

*Remark:* Figures 24–29 contain all possibilities, modulo rotations and reflections, for obtaining the sum $26$.