Your solution for $(1)$ is correct, and you can apply the same ideas to $(2)$.
First choose $3$ nickels, then choose the remaining $7$ coins, in $\binom{4+7-1}7=\binom{10}7=120$ ways. But there should be no more than $2$ quarters, so we need to subtract the choices with more than $2$ quarters. First choose $3$ nickels and $3$ quarters, then choose the remaining $4$ coins, in $\binom{4+4-1}4=\binom74=35$ ways. So the desired number is $120-35=85$.