Checking if u[n] and v[n] are orthogonal
It's essentially asking you to show that two signals that are orthogonal in the baseband (their original form) remain orthogonal after amplitude modulation process. In other words, amplitude modulation preserves orthogonality.

Think about the question this way. You have two signals that are orthogonal, meaning that

$$ \sum_{m=-\infty}^{\infty}x[m]y[m] = 0 $$

This is a bit ugly, since we have to deal with infinities. However, your problem also says that they both have bandwidth $B$, which means that in frequency domain they will have finite width.

Let's define $\mathscr{B}$ as

$$ \mathscr{B} = \frac{B}{F_s}$$

which means that in your discrete Fourier transform signals $X[k]$ and $Y[k]$, the only non-zero samples exist for $-\frac{\mathscr{B}}{2} \le k \le \frac{\mathscr{B}}{2}$. Since DFT is an orthogonal transform, signals orthogonal in time domain will remain orthogonal in frequency domain, i.e.

$$ \sum_{l=-\infty}^{\infty}X[l]Y[l] =\sum_{l=-\mathscr{B}/2}^{\mathscr{B}/2}X[l]Y[l] = \sum_{m=-\infty}^{\infty}x[m]y[m] = 0 $$

We know that multiplying a signal by a cosine shifts is away from DC in frequency domain, and the fact that the modulating frequency $f_c$ is larger than $B$ means that this won't cause any problems with overlapping modulated bands (think about what would happen if this weren't the case).

I think this gives you enough hits to take it from here.