, 2016

The answer is yes for $75\%$ and no for $60\%$.
Let the FRT was less than $75\%$ after the first half but eventually greater than $75\%$. Then there is a successful free throw in the second half such that after it the FRT became at least $75\%$. Consider the first such free throw $S$. We claim that after $S$ the FRT has become exactly $75\%$. Let the FRT before $S$ be $\frac{x}{y}$ where $y$ is the total number of free throws before $S$ and $x$ is the number of successful ones among them. Then the FRT after $S$ is $\frac{x+1}{y+1}$, and by assumption, $\frac{x}{y} < \frac{3}{4} \leq \frac{x+1}{y+1}$. The left inequality gives $4x < 3y$, the right one yields $3y \leq 4x+1$. Hence $4x < 3y \leq 4x+1$, and because $3y$ is an integer, it follows that $3y = 4x+1$. It is immediate that this equality is equivalent to $\frac{x+1}{y+1} = \frac{3}{4}$.
Therefore $S$ made the FRT exactly $75\%$.

The case $60\%$ is different. Let Mateo score $4$ free throws out of a total of $7$ in the first half. Then his FRT after the first half is $\frac{4}{7} < \frac{3}{5}$. Suppose also that all of his free throws in the second half are successful. The first one of them makes the FRT equal to $\frac{5}{8} > \frac{3}{5}$. Each subsequent free throw, being successful, increases the FRT (because if $0 < u < v$ then $\frac{u}{v} < \frac{u+1}{v+1}$). So the FRT will be greater than $60\%$ at the end of the game, but never exactly equal to $60\%$ throughout. Naturally there are (infinitely) many fractions that can replace $\frac{4}{7}$ in this argument: $\frac{1}{2}$, $\frac{7}{12}$, $\frac{10}{17}$, $\frac{13}{22}$, $\frac{16}{27}$ etc.