= Solution
For $t>0$, put $M_t=\sup_{0\leq s\leq t}B_s$. For $\lambda>0$, reflect the Brownian path after its first hit of $\lambda$. The <Strong Markov property> and symmetry of subsequent <Brownian increments> make this reflection preserve the path distribution. It exchanges the events $\{M_t\geq\lambda,B_t<\lambda\}$ and $\{B_t>\lambda\}$. Since $B_t$ has no atom at $\lambda$, the <Brownian reflection principle> gives
$$
\boxed{\mathbb P(M_t>\lambda)=2\mathbb P(B_t>\lambda)=2\bigl(1-\Phi(\lambda/\sqrt t)\bigr),\qquad t>0,\ \lambda\geq0.}
$$
The case $\lambda=0$ follows by decreasing positive levels to zero: the <probability> is one. At $t=0$, $M_0=0$ and the <probability> is zero for every $\lambda\geq0$.
The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the <half-normal distribution> of $\sqrt t\,|Z|$ with $Z\sim N(0,1)$. Thus
$$
\boxed{\|M_t\|_2=\sqrt{\mathbb E B_t^2}=\sqrt t,\qquad t\geq0.}
$$
For $A_t=\sup_{s\leq t}|B_s|$, the pointwise inequality $A_t\geq M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda>0$. Reflection at $\lambda$ also gives
$$
\mathbb P(B_t<-\lambda,M_t\geq\lambda)=\mathbb P(B_t>3\lambda).
$$
Subtracting this from $\mathbb P(B_t<-\lambda)=\mathbb P(B_t>\lambda)$ gives
$$
\mathbb P(B_t<-\lambda,M_t<\lambda)
=\mathbb P(B_t>\lambda)-\mathbb P(B_t>3\lambda)>0.
$$
On this event $A_t\geq|B_t|>\lambda>M_t$, so $A_t^2-M_t^2$ is strictly positive with positive <probability>. Finally, $|B_t|$ is a continuous nonnegative <submartingale> by the conditional <Jensen inequality>, and part (a) gives $\|A_t\|_2\leq2\|B_t\|_2=2\sqrt t$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event gives
$$
\boxed{\sqrt t<\|A_t\|_2\leq2\sqrt t\quad(t>0).}
$$
This is the <strict comparison of one-sided and absolute Brownian maxima>. The strict lower inequality printed with $t\geq0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.
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