How many different lottery draws are possible in which the 5 drawn numbers form a) an arithmetic, b) a geometric sequence?
a) Let the drawn numbers be $a, a+d, a+2d, a+3d$, and $a+4d$. Since an arithmetic sequence is uniquely determined by its first term and common difference, the above quintuples of numbers can be uniquely associated with pairs of positive integers $(a, d)$, where $a+4d \leq 90$. Therefore, our task can be restated as follows: how many pairs of positive integers $(a, d)$ exist such that $a+4d \leq 90$?

Fix the value of $d$! Since $a+4d \leq 90$, we have $a \leq 90-4d$, so the possible values of $a$ are $1, 2, 3, \ldots, 90-4d$. (If $d \geq 23$, then there are no suitable $(a, d)$ pairs.) Therefore, the number of ordered pairs $(a, d)$ that satisfy the conditions of the problem is

$$
\sum_{d=1}^{22}(90-4d)=\frac{22 \cdot(90-4 \cdot 1+90-4 \cdot 22)}{2}=968
$$

b) If the drawn numbers form a geometric sequence, let the smallest number be $a$ and the common ratio of the sequence be $q$. Then the drawn numbers are $a, aq, aq^2, aq^3, aq^4$, where $q>1$ and $q$ is rational, since it can be expressed as the ratio of two integers (for example, $aq$ and $a$). Let

$$
q=\frac{p}{r}
$$

where $p$ and $r$ are relatively prime positive integers. We distinguish three cases based on the value of $r$:

1. $r=1$, i.e., $q$ is an integer. In this case, since $a \cdot q^4 \leq 90$, we have $q^4 \leq 90$, so $q \leq 3$. If $q=2$, then $a \leq 5$, giving us the solutions $(1, 2, 4, 8, 16)$, $(2, 4, 8, 16, 32)$, and $(3, 6, 12, 24, 48)$. If $q=3$, then $a \leq 1$, giving us the solution $(1, 3, 9, 27, 81)$. If $q \geq 4$, then $a \cdot q^4 \geq 256$, so there are no more solutions.
2. $r=2$. In this case, $a \cdot q^4 = a \cdot \left(\frac{p}{2}\right)^4$ must be an integer, so $a$ must be a multiple of $16$. Since $a \cdot q^4 \leq 90$, we have $a \cdot \left(\frac{p}{2}\right)^4 \leq 90$. If $p \geq 3$, then $a \cdot q^4 \geq 16 \cdot \left(\frac{3}{2}\right)^4 = 81$. For $a=16$ and $p=3$, we get the solution $(16, 24, 36, 54, 81)$. If $a > 16$ or $p > 3$, then

$$
a q^4 \geq 32 \cdot \left(\frac{3}{2}\right)^4 = 162, \text{ or } \quad a q^4 \geq 16 \cdot \left(\frac{4}{2}\right)^4 = 256
$$

so there are no more solutions.
3. $r \geq 3$. In this case, similarly to the previous cases, from the fact that $a q^4 = a \frac{p^4}{r^4}$ is an integer, it follows that $a \geq r^4$. On the other hand, since $p > r$, we have $p \geq r+1$, and thus

$$
a q^4 \geq r^4 \cdot \left(\frac{r+1}{r}\right)^4 = (r+1)^4 \geq 4^4 = 256
$$

so there are no solutions in this case.

Thus, we have a total of 7 solutions.