Question:

<div class='solution'><p><strong>Step 1:</strong> $f(xy)=f(x)f(y)$ with $f(0)\ne0$. Set $x=y=0$: $f(0)=f(0)^2$ → $f(0)=1$ (since $\ne0$). Set $y=1$: $f(x)=f(x)f(1)$ → $f(1)=1$. For continuous solutions: $f(x)=1$ for all $x\in[0,1]$ (only solution with $f(0)=f(1)=1$ and multiplicative property).</p><p><strong>Step 2:</strong> $y'=1$, $y(0)=1$ → $y=x+1$.</p><p><strong>Step 3:</strong> $y(1/4)+y(3/4)=(1/4+1)+(3/4+1)=5/4+7/4=3$. <strong>Answer: (1)</strong> $3$.</p></div>