1. **Problem statement:** We have a function $f$ defined for all real numbers such that $$f(x+y) = f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right)$$ for all real $x,y$. We are given that $$f(8) + f(2) = 6$$ and need to find $f(128)$. 2. **Analyze the functional equation:** Substitute $y=0$ to get $$f(x) = f\left(\frac{x}{2}\right) + f(0)$$ for all $x$. Rearranged, this is $$f(x) - f\left(\frac{x}{2}\right) = f(0).$$ 3. **Define $c = f(0)$** and rewrite as $$f(x) = f\left(\frac{x}{2}\right) + c.$$ 4. **Iterate the relation:** Applying repeatedly, $$f(x) = f\left(\frac{x}{2}\right) + c = f\left(\frac{x}{4}\right) + 2c = \cdots = f\left(\frac{x}{2^n}\right) + nc.$$ 5. **Take limit as $n \to \infty$:** Since $f$ is defined on all reals, assume $f$ is bounded near zero and $\lim_{n\to\infty} f\left(\frac{x}{2^n}\right) = f(0) = c$. Then $$f(x) = c + nc.$$ But this grows without bound unless $c=0$. So $c=0$ and $$f(x) = f\left(\frac{x}{2}\right).$$ 6. **From $f(x) = f\left(\frac{x}{2}\right)$ for all $x$,** it follows that $f$ is constant on all positive numbers (and by extension all reals). Let $f(x) = k$ for some constant $k$. 7. **Use the given condition:** $$f(8) + f(2) = k + k = 2k = 6 \implies k = 3.$$ 8. **Find $f(128)$:** Since $f$ is constant, $$f(128) = k = 3.$$ 9. **Check answer choices:** None of the options (4, 8, 16, 32) equals 3. This suggests a reconsideration. 10. **Re-examine the original equation:** Try $x=y=0$: $$f(0) = f(0) + f(0) \implies f(0) = 2f(0) \implies f(0) = 0.$$ So $c=0$ confirmed. 11. **Try $y=x$:** $$f(2x) = f\left(\frac{x}{2}\right) + f\left(\frac{x}{2}\right) = 2f\left(\frac{x}{2}\right).$$ 12. **Rewrite as:** $$f(2x) = 2f\left(\frac{x}{2}\right).$$ Substitute $x$ by $4x$: $$f(8x) = 2f(2x).$$ 13. **From step 11, $f(2x) = 2f\left(\frac{x}{2}\right)$, so:** $$f(8x) = 2f(2x) = 2 \times 2f\left(\frac{x}{2}\right) = 4f\left(\frac{x}{2}\right).$$ 14. **Try a power function form:** Suppose $f(x) = A x^m$. Then $$f(x+y) = A(x+y)^m,$$ and $$f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right) = A\left(\frac{x}{2}\right)^m + A\left(\frac{y}{2}\right)^m = A 2^{-m} (x^m + y^m).$$ 15. **Equate:** $$A(x+y)^m = A 2^{-m} (x^m + y^m).$$ Divide both sides by $A$ (assuming $A \neq 0$): $$(x+y)^m = 2^{-m} (x^m + y^m).$$ 16. **Test $m=1$:** $$(x+y)^1 = 2^{-1} (x + y) \implies x + y = \frac{x + y}{2},$$ which is false unless $x+y=0$. 17. **Test $m=0$:** $$(x+y)^0 = 2^{0} (x^0 + y^0) \implies 1 = 1 (1 + 1) = 2,$$ false. 18. **Try $m=-1$:** $$(x+y)^{-1} = 2^{1} (x^{-1} + y^{-1}) = 2 \left(\frac{1}{x} + \frac{1}{y}\right).$$ This is not generally true. 19. **Try $m=2$:** $$(x+y)^2 = 2^{-2} (x^2 + y^2) \implies (x+y)^2 = \frac{1}{4} (x^2 + y^2).$$ This is false for general $x,y$. 20. **Try $f(x) = kx$ linear:** Then $$f(x+y) = k(x+y),$$ and $$f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right) = k\frac{x}{2} + k\frac{y}{2} = \frac{k}{2}(x+y).$$ 21. **Equate:** $$k(x+y) = \frac{k}{2}(x+y) \implies k = \frac{k}{2} \implies k=0.$$ So $f(x) = 0$ is a solution. 22. **Check given condition:** $$f(8) + f(2) = 0 + 0 = 0 \neq 6.$$ So $f(x) = 0$ is not the solution. 23. **Try $f(x) = c$ constant:** Then $$f(x+y) = c,$$ and $$f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right) = c + c = 2c.$$ 24. **Equate:** $$c = 2c \implies c=0.$$ So $f(x) = 0$ again. 25. **Try $f(x) = A$ for $x=0$ and $f(x) = B$ for $x \neq 0$:** This is complicated and unlikely. 26. **Try $x=y$ in original:** $$f(2x) = 2 f\left(\frac{x}{2}\right).$$ Let $g(x) = f(2x)$, then $$g(x) = 2 f\left(\frac{x}{2}\right) = 2 f\left(\frac{x}{2}\right).$$ This is circular. 27. **Try $x=2a$, $y=2b$:** $$f(2a + 2b) = f(a) + f(b).$$ Let $h(t) = f(2t)$, then $$h(a+b) = f(a) + f(b).$$ 28. **Rewrite:** $$f(a) + f(b) = h(a+b) = f(2(a+b)).$$ 29. **Set $a=b=0$:** $$f(0) + f(0) = f(0) \implies 2f(0) = f(0) \implies f(0) = 0.$$ 30. **From step 28:** $$f(a) + f(b) = f(2a + 2b).$$ Let $x=a$, $y=b$, then $$f(x) + f(y) = f(2x + 2y).$$ 31. **Set $y=0$:** $$f(x) + f(0) = f(2x) \implies f(x) = f(2x)$$ since $f(0)=0$. 32. **From step 31:** $$f(x) = f(2x) = f(4x) = \cdots$$ so $f$ is constant on all nonzero multiples of $x$. 33. **Use $f(8) + f(2) = 6$:** Since $f(8) = f(2)$, $$2 f(2) = 6 \implies f(2) = 3.$$ 34. **Since $f(x) = f(2x)$, $f(128) = f(64) = f(32) = f(16) = f(8) = f(4) = f(2) = 3.$** 35. **Answer:** $f(128) = 3$. 36. **Check options:** None match 3, so the problem likely expects the value $f(128) = 3$ which is not listed. Possibly a typo or misprint in options. **Final conclusion:** $\boxed{3}$ is the value of $f(128)$ based on the functional equation and given condition.