To get to $59$ cents, you know that at least $4$ of the $38$ coins must be pennies. This brings the known counts to one quarter, one dime, one nickel, and $5$ pennies, and leaves $34$ unassigned coins to total $55$ cents. At most $11$ of these can be nickels or higher, so at least $23$ must be pennies. Since pennies must come in groups of $5$, at least $25$ must be pennies. This brings the known counts to one quarter, one dime, one nickel, and $30$ pennies, and leaves $9$ coins to total $30$ cents. Once again, at most $6$ of these coins can be a nickel or higher, so at least $3$ must be pennies, which means, again, that at least $5$ must be pennies. This brings the known counts to one quarter, one dime, one nickel, and $35$ pennies, and leaves $4$ coins to total $25$ cents.
At this point, pennies can no longer be used (since they must come in groups of $5$), nor can quarters. It's not hard to jump straight to the answer, but let's be methodical: Since $25\gt4\cdot5$, we need at least one dime, and since $25\lt3\cdot10$, we can use at most two dimes, so we need at least $2$ nickels. This leaves $1$ coin to total $5$ cents, i.e. another nickel, so the $4$ coins totalling $25$ cents must be $3$ nickels and one dime. This brings to final known counts to $1$ quarter, $2$ dimes, $4$ nickels, and $35$ pennies.
Alternatively, let $P=p+1$, $N=n+1$, $D=d+1$, and $Q=q+1$ be the number of pennies, nickels, dimes, and quarters, where $p,n,d,q\ge0$. We have $P+N+D+Q=42$ and $P+5N+10D+25Q=100$, which imply
Subtracting these gives
which immediately implies $q=0$, leaving $4n+9d=21$. Since $21$ is odd while $4$ is even, we must have $d$ odd, which means $d=1$ (since $d\ge3\implies9d\ge27\gt21$) and thus $4n=12$, so $n=3$. Thus $p=38-3-1-0=34$, hence $(P,N,D,Q)=(35,4,2,1)$.