 Solution

For an odd composite $N$ and a unit $b$ modulo $N$, $N$ is a Fermat pseudoprime to base $b$ whenIt is an Euler pseudoprime to base $b$ whenwhere the right side is the Jacobi symbol.

By the Chinese remainder theorem,and $\phi (105)=2\cdot 4\cdot 6=48$. The condition $b^{104}=1$ is automatic modulo $3$ and $5$. Modulo $7$ it hassolutions. Thus there are $2\cdot 4\cdot 2=16$ Fermat bases, giving proportion

For the Euler condition, $b^{52}=1$ modulo $3$, so the Jacobi symbol must be $+1$. The power condition is automatic modulo $5$, while modulo $7$ it again restricts $b$ to the two solutions of $b^2=1$. For each of those two residues, exactly half of the $2\cdot 4$ choices modulo $3$ and $5$ have Jacobi symbol $+1$. Hence there are $2\cdot 4=8$ Euler bases and proportion