Tree diagrams and conditional probability

The branches of a tree carry PROBABILITIES (fractions or decimals), not whole-number counts. Writing $3$ and $2$ from "3 red, 2 blue" is wrong — write $\tfrac35$ and $\tfrac25$. Also give each branch to the accuracy the question asks for.

Without replacement, the second pick has a smaller total, so the denominators must change. Treating it like "with replacement" (same denominators on the second set) scores nothing. From $5$ counters, a second pick is out of $4$, e.g. $\tfrac35\times\tfrac24$, not $\tfrac35\times\tfrac35$.

For "one red and one blue" there are TWO paths — red-then-blue and blue-then-red — and you must add both. Counting only one order halves the answer, e.g. giving $\tfrac{6}{20}$ instead of $\tfrac{6}{20}+\tfrac{6}{20}=\tfrac{12}{20}=\tfrac35$.

Along a single path you MULTIPLY the branch probabilities; adding them instead can give a value above $1$, which is impossible. Add only BETWEEN separate complete paths. If a path probability comes out greater than $1$, you have added where you should have multiplied.

Without replacement, BOTH numbers change on the second pick: the total drops by $1$ AND the count of the colour just taken drops by $1$. After taking one red from $3$ red of $5$, the next red branch is $\tfrac24$ — not $\tfrac34$ (numerator forgotten) and not $\tfrac35$ (nothing changed).

For "at least one", adding up every favourable path is slow and error-prone. Use the complement instead: $P(\text{at least one})=1-P(\text{none})$. E.g. $P(\text{at least one blue})=1-P(\text{both red})=1-\tfrac35\times\tfrac24=\tfrac{7}{10}$.