It seems that the claim is false in general since problems arise when there's a non-null probability of hitting a shell. What follows is a counter-example (at least if I haven't made any mistake).
Let $m=2$ and define
\begin{equation*}
\sigma^1:[0,+\infty)^2\to\{1,2\}, (r_1,r_2)\mapsto \min\left({\operatorname{argmin}_{k\in\{1,2\}}}r_k\right)
\end{equation*}
so that
\begin{equation*}
\forall r_1,r_2\ge 0, \sigma ^1(r_1,r_2)=
\begin{cases}
1, &\text{if $r_1 \le r_2$;}\\
2, &\text{otherwise;}\\
\end{cases}
\end{equation*}
and
\begin{equation*}
\forall r_1,r_2\ge 0, \sigma ^2(r_1,r_2)=
\begin{cases}
2, &\text{if $r_1 \le r_2$;}\\
1, &\text{otherwise.}\\
\end{cases}
\end{equation*}
Define $\mathcal{X}=\{0,1\}$ and let $d$ be the discrete metric on $\mathcal{X}$.
Let $p_0\in (0,1)$ and let $\mathbb{P}_X$ be the unique probability measure on $2^{\{0,1\}}$ such that $\mathbb{P}_{X}(\{0\})=p_0$. Let $x=0$ and $A=\{x\}$.
Let's show that
\begin{equation*}
\mathbb{P}\left(X^1 \in A | W=1\right) \neq \mathbb{P}\left(X \in A | d(x,X)\le 1\right)
\end{equation*}
First, notice that
\begin{equation*}
\mathbb{P}\left(X \in A | d(x,X)\le 1\right) = \mathbb{P}\left(X \in A\right) = \mathbb{P}(X=0) = p_0.
\end{equation*}
On the other hand
\begin{equation*}
\{W=1\} = \{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)
\end{equation*}
and
\begin{align*}
\{X^1\in A\} \cap \{W=1\} &= \left(\{X^1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X^1=0\}\cap\{X_1 = 1\}\cap \{X_2 = 0\}\right)\\
&= \left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right),\\
\end{align*}
so
\begin{align*}
\mathbb{P}\left(X^1\in A | W=1\right) &= \frac{\mathbb{P}\left(\{X^1\in A\}\cap \{W=1\}\right)}{\mathbb{P}\left(\{W=1\}\right)}\\
&= \frac{\mathbb{P}\left(\left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}{\mathbb{P}\left(\{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}\\
&= \frac{2 p_0(1-p_0)}{1-p_0+p_0(1-p_0)}= \frac{2 p_0(1-p_0)}{1-p_0^2}\\
&= \frac{2 p_0}{1+p_0}\neq p_0=\mathbb{P}\left(X \in A | d(x,X)\le 1\right).
\end{align*}
I agree to your solution of b). Let me define the events:
D: Difficult exam, $\overline D:$ Easy exam
C: First question is marked as challanging, $\overline C:$ First question is not marked as challanging
For a) you use the law of total probability:
$$P( C)=P(D\cap C)+P(\overline D\cap C),$$
where $P(D\cap C)=P(D)\cdot P(C|D)$ and $P(\overline D\cap C)=P(\overline D)\cdot P(C|\overline D)$
Can you proceed?
Best Answer
$\color{purple}{\checkmark}$ Solution verified.
Yeah, it looks good. You might want to include steps to indicate where you drew the numbers though.
... $\mathbb P(\text{dead}) = \mathbb P(\text{dead}\mid\text{watered})\cdot\mathbb P(\text{watered})+\mathbb P(\text{dead}\mid\text{forgot})\cdot\mathbb P(\text{forgot})$ et cetera.