Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the same number is rolled six times. Thus, the probability of exactly two numbers appearing in six rolls of the die is
$$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$
Exactly three numbers appear:
There are $\binom{6}{3}$ ways for three of the six numbers to appear. On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers. However, we have counted the sequences in which not all three of the numbers appear. There are $\binom{3}{2}$ ways for two of the numbers to appear. For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times. Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear. The number of sequences in which exactly one of the three numbers appears is $3$. Hence, the probability that exactly three of the numbers appear is
$$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$
My thanks to @bof for clarifying my thinking about the first problem. Any errors that remain are entirely my responsibility.
First die can be 2, 4, or 6. Second any roll 1-6. So the sample space is $3 \cdot 6= 18$
Only two positive outcomes: {2,4} and {4,2}.
Probability $= \frac{2}{18} = \frac{1}{9}$
Best Answer
Record the result of the tossing as an ordered pair $(a,b)$, where $a$ and $b$ are integers between $1$ and $4$. Here the first coordinate records the result of the first toss, and the second coordinate records the result of the second toss.
(a) Let the number of such ordered pairs be $N$. What is the value of $N$? There is a fast way of finding $N$. But one can do it slowly, by listing and counting. Here is a start: $(1,1)$, $(1,2)$, $(1,3)$, $(1,4)$.
(b) All these ordered pairs are equally likely. How many ordered pairs give you a sum of $4$ or less? List and count them carefully. Let the number be $S$.
(c) Your probability is $\dfrac{S}{N}$.