For part (b), I found the CDF to be $(1-x)^n$ and I differentiated that to get the PDF of $-n(1-x)^{n-1}$.
I'm quite confused on how to calculate the Expectation of M.
My current workings are as such :
E(M) = E$\left(\frac{X_1+...+X_{10}}{10}\right)$ = $\frac 1{10}$(E$X_1$ + ... + E$X_{10}$) = $\frac 1{10}$ x 10(E$X_1$) = ...
Since all the X are uniformly distributed, my idea is that I would be able to simplify it as above but I do not know how to continue from there.
Any help would be appreciated thanks~
The c.d.f. of $M$ is
$$ F_M(x) = \begin{cases} 0, & x \leq 0 \\ x^{10}, & 0 < x < 1 \\ 1, & x \geq 1 \end{cases} $$
Differentiating this gives $f_M(x) = 10 x^9$ on $(0, 1)$ and zero outside.
I am not sure what led you to believe that $\mathbf{E}[M] = \mathbf{E}[\frac{1}{10}(X_1+\cdots+X_{10})]$, but this is simply not true. Since you know the p.d.f. of $M$, you can compute its expectation using the integral
$$ \mathbf{E}[M] = \int_{-\infty}^{\infty} x f_M(x) \, dx = \int_{0}^{1} 10x^{10} \, dx = \frac{10}{11}. $$
Of course, you can compute $\mathbf{E}[X_1]$ using the same idea, leading to
$$ \mathbf{E}[X_1] = \int_{-\infty}^{\infty} x f_{X_1}(x) \, dx = \int_{0}^{1} x \, dx = \frac{1}{2}. $$