Let $X=\max\{B_t: 0 \le t \le 1\}$ and set $Y=B_2-B_1$
(i) Prove that $P(X>Y)\ge1/2$.
(ii) Calculate the distribution functions of $X$ and of $Z=|Y|$.
(iii) Prove or disprove that $P(X>|Y|)\ge 1/2$.
While I don't know how to solve (i), I know that $f_Z(t)=2*f_{N(0,1)}(t)$, $\forall t \ge0$ and zero else. For this reason, I'm inclined to think that (iii) is incorrect.
By the reflection principle you get $P(X>a)=2P(B_1>a)$ for $a>0$. Thus the distribution of $X$ is given by $$P(X\leq a)=1-2P(B_1>a)=1-2\int_a^\infty\frac{\exp(-\frac{x^2}{2})}{\sqrt{2\pi}}\mathrm dx $$ and the pdf is $f(x)=\sqrt{\frac{2}{\pi}}\exp(-\frac{x^2}{2})\cdot1_{]0,\infty]}(x)$. Now you know the pdf of two independet random variables $X$ and $Y$, so you can use the convolution of probability distributions to calculate the pdf of $X-Y=X+(-Y)$ (note: $Y$ and $-Y$ have the same distribution). Then you just have to calculate the integral $$P(X-Y>0)=\int_0^\infty\int_0^\infty\sqrt{\frac{2}{\pi}}\exp\left(-\frac{x^2}{2}\right)\frac{\exp(-\frac{(z-x)^2}{2})}{\sqrt{2\pi}}\mathrm dx\mathrm dz. $$ This is the solution for (i) and the first part of (ii). The second part of (ii) is already solved by you and (iii) can solved in the same way as above by considering $P(X-|Y|>0)$.