For problem 11 , i couldn’t find the correct translation , so i just posted the hungarian version . If anyone could translate it ,i would be very thankful .
[tip=see hungarian]Az $X$  ́es$ Y$ valo ́s  ́ert ́eku ̋ v ́eletlen v ́altoz ́ok maxim ́alkorrel ́acio ́ja az $f(X)$  ́es $g(Y )$ v ́altoz ́ok korrela ́cio ́j ́anak szupr ́emuma az olyan $f$  ́es $g$ Borel m ́erheto ̋, $\mathbb{R} \to \mathbb{R}$ fu ̈ggv ́enyeken, amelyekre $f(X)$  ́es $g(Y)$ v ́eges sz ́ora ́su ́. Legyen U a $[0,2\pi]$ interval- lumon egyenletes eloszl ́asu ́ val ́osz ́ınu ̋s ́egi v ́altozo ́, valamint n  ́es m pozit ́ıv eg ́eszek. Sz ́am ́ıtsuk ki $\sin(nU)$  ́es $\sin(mU)$ maxim ́alkorrela ́ci ́oja ́t. [/tip]
Edit:
[hide=Translation thanks to @tintarn] The maximal correlation of two random variables $X$ and $Y$ is defined to be the supremum of the correlations of $f(X)$ and $g(Y)$ where $f,g:\mathbb{R} \to \mathbb{R}$ are measurable functions such that $f(X)$ and $g(Y)$ is (almost surely?) finite.
Let $U$ be the uniformly distributed random variable on $[0,2\pi]$ and let $m,n$ be positive integers. Compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$.
(Remark: It seems that to make sense we should require that $E[f(X)]$ and $E[g(Y)]$ as well as $E[f(X)^2]$ and $E[g(Y)^2]$ are finite.
In fact, we may then w.l.o.g. assume that $E[f(X)]=E[g(Y)]=0$ and $E[f(Y)^2]=E[g(Y)^2]=1$.)[/hide]
1. **Understanding the Problem:**
   The problem asks us to compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$, where $U$ is a uniformly distributed random variable on $[0, 2\pi]$, and $m$ and $n$ are positive integers. The maximal correlation of two random variables $X$ and $Y$ is defined as the supremum of the correlations of $f(X)$ and $g(Y)$ over all measurable functions $f$ and $g$ such that $f(X)$ and $g(Y)$ are finite.

2. **Correlation Definition:**
   The correlation $\rho$ between two random variables $A$ and $B$ is given by:
   \[
   \rho(A, B) = \frac{\text{Cov}(A, B)}{\sqrt{\text{Var}(A) \text{Var}(B)}}
   \]
   where $\text{Cov}(A, B)$ is the covariance of $A$ and $B$, and $\text{Var}(A)$ and $\text{Var}(B)$ are the variances of $A$ and $B$, respectively.

3. **Uniform Distribution Properties:**
   Since $U$ is uniformly distributed on $[0, 2\pi]$, its probability density function is:
   \[
   f_U(u) = \frac{1}{2\pi}, \quad 0 \leq u \leq 2\pi
   \]

4. **Expectation and Variance Calculations:**
   We need to calculate the expectations and variances of $\sin(nU)$ and $\sin(mU)$:
   \[
   E[\sin(nU)] = \int_0^{2\pi} \sin(nu) \cdot \frac{1}{2\pi} \, du = 0
   \]
   because the integral of $\sin(nu)$ over a full period is zero.

   Similarly,
   \[
   E[\sin(mU)] = 0
   \]

   The variances are:
   \[
   \text{Var}(\sin(nU)) = E[\sin^2(nU)] - (E[\sin(nU)])^2 = E[\sin^2(nU)]
   \]
   \[
   E[\sin^2(nU)] = \int_0^{2\pi} \sin^2(nu) \cdot \frac{1}{2\pi} \, du = \frac{1}{2}
   \]
   because $\sin^2(x)$ has an average value of $\frac{1}{2}$ over a full period.

   Thus,
   \[
   \text{Var}(\sin(nU)) = \frac{1}{2}
   \]
   and similarly,
   \[
   \text{Var}(\sin(mU)) = \frac{1}{2}
   \]

5. **Covariance Calculation:**
   The covariance $\text{Cov}(\sin(nU), \sin(mU))$ is:
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)] - E[\sin(nU)]E[\sin(mU)]
   \]
   Since $E[\sin(nU)] = 0$ and $E[\sin(mU)] = 0$, we have:
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)]
   \]

   Using the product-to-sum identities:
   \[
   \sin(nU) \sin(mU) = \frac{1}{2} [\cos((n-m)U) - \cos((n+m)U)]
   \]

   Therefore,
   \[
   E[\sin(nU) \sin(mU)] = \frac{1}{2} \left( E[\cos((n-m)U)] - E[\cos((n+m)U)] \right)
   \]

   Since $U$ is uniformly distributed over $[0, 2\pi]$, the expectations of $\cos((n-m)U)$ and $\cos((n+m)U)$ are zero:
   \[
   E[\cos((n-m)U)] = 0 \quad \text{and} \quad E[\cos((n+m)U)] = 0
   \]

   Thus,
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = 0
   \]

6. **Correlation Calculation:**
   Finally, the correlation $\rho(\sin(nU), \sin(mU))$ is:
   \[
   \rho(\sin(nU), \sin(mU)) = \frac{\text{Cov}(\sin(nU), \sin(mU))}{\sqrt{\text{Var}(\sin(nU)) \text{Var}(\sin(mU))}} = \frac{0}{\sqrt{\frac{1}{2} \cdot \frac{1}{2}}} = 0
   \]

7. **Maximal Correlation:**
   Since the correlation is zero, the maximal correlation is also zero.

\[
\boxed{0}
\]