(a) We know that $$P(x)=\binom{x}{0}+\binom{x}{1}+ \dots +\binom{x}{n}$$satisfies $P(i)=2^i$ for $i=0,1, \dots, n$. Also, it must be the only polynomial of degree $n$ taking these values at $0,1, \dots, n$. So, if Professor D made no mistake, then this must be his polynomial. But then $P(n+1)=2^{n+1}-1$, a contradiction. (b) For the above polynomial $P$, we have $P(n+2)=2^{n+2}-n-3$, and so the minimum number of mistakes made is simply $1$. We just need to change the number on office $n+1$ from $2^{n+1}$ to $2^{n+1}-1$.

Same as @above; posting for storage. (a) We claim the polynomial is $f(x)=\sum_{j=0}^n\binom{x}{j}.$ Indeed, $f(i)=2^i$ for $i=0,1,\dots,n$ so the identity theorem implies the desired conclusion. Note $$f(n+1)=\sum_{j=0}^{n+1}\binom{n+1}{i}=2^{n+1}-1\neq 2^{n+1},$$a contradiction. (b) Notice $f(n+2)=\sum_{j=0}^{n+2}\binom{n+2}{i}=2^{n+2}-(n+2)-1=2^{n+2}-n-3,$ so the professor did not make a mistake for $f(n+2).$ Hence, the minimum amount of errors he made is one, at $f(n+1).$ To correct the error, set $f(n+1)=2^{n+1}-1.$ $\square$

(a) We will show that such a polynomial cannot exists. Note that we can write $$P(x)=\sum_{i=0}^n 2^i \left(\prod_{j=0}^{i-1} \frac{x-j}{i-j}\right)\left(\prod_{j=i+1}^n \frac{x-j}{i-j}\right)\qquad (\star)$$since $\deg P=n$ and LHS and RHS agree on $n+1$ values, namely $x=0,1,\dots,n$. So putting $x=n+1$ we get \begin{align*} P(n+1) &= \sum_{i=0}^n 2^i \left(\prod_{j=0}^{i-1} \frac{n+1-j}{i-j}\right)\left(\prod_{j=i+1}^n \frac{n+1-j}{i-j}\right)\\&= \sum_{i=0}^n 2^i \cdot \frac{(n+1)!}{(n+1-i)!i!}\cdot \frac{(n-i)!}{(-1)^{n-i}(n-i)!} \\ &=\sum_{i=0}^n 2^i (-1)^{n-i} \binom{n+2}i \\ &=-\sum_{i=0}^{n+1} 2^i (-1)^{n+1-i}\binom{n+1}i -2^{n+1}(-1)^{n-n-1}\binom{n+1}{n+1} \\ &=-(2-1)^{n+1}+2^{n+1}=2^{n+1}-1\end{align*}this is a contradiction. (b) I will show that we only need to change one value to fix the error. In particular, I'll change the value of $P(n+1)$. By $(\star)$ we find \begin{align*} P(n+2) &=\sum_{i=0}^n 2^i \prod_{j=0}^{i-1} \frac{n+2-j}{i-j} \prod_{j=i+1}^n \frac{n+2-j}{i-j} \\ &=\sum_{i=0}^n 2^i \frac{(n+2)!}{(n+2-i)! i!} \frac{(n+1-i)!}{(-1)^{n-i}(n-i)!} \\ &=\sum_{i=0}^n 2^i (-1)^{n-i} (n+1-i)\binom{n+2}i \\ &=2^n \sum_{i=0}^n (-1/2)^{n-i}(n+1-i)\binom{n+2}{i}+2^{n+2} \\ &=2^{n+2}+2^n \lim_{x\to-1/2}\frac{d}{dx} \sum_{i=0}^n x^{n+1-i}\binom{n+2}i \\ &= 2^{n+2}+2^n \lim_{x \to -1/2} \frac{d}{dx} (1/x \cdot (1+x)^{n+2}) \\ &=2^{n+2}-n-3\end{align*}Thus, according to (a), changing $P(n+1)=2^{n+1}-1$ and letting everything else be as it is will work. Now we will show that changing any value other than $P(n+1)$ will not work. Obviously changing $P(n+2)$ won't do anything (by (a)). Changing $P(0)$ doesn't work because then consider $$P(x)=\sum_{i=0}^n 2^{i+1} \prod_{j=0}^{i-1} \frac{x-j-1}{i-j}\prod_{j=i+1}^n \frac{x-j-1}{i-j}$$(since LHS and RHS agree on the $n+1$ points $x=1,\dots,n+1$). So, $$P(n+2)=2\sum_{i=0}^n 2^i \left(\prod_{j=0}^{i-1} \frac{n+1-j}{i-j}\right)\left(\prod_{j=i+1}^n \frac{n+1-j}{i-j}\right)=2(2^{n+1}-1)=2^{n+2}-2$$contradiction. Changing $P(k)$ (where $k\neq 0,n+1$) doesn't work. As before write $$P(x)=\sum_{i=0}^{k-1} 2^i \prod_{j=0}^{i-1} \frac{x-a_j}{i-a_j} \prod_{j=i+1}^n \frac{x-a_j}{i -a_j}+\sum_{i=k}^n 2^{i +1} \prod_{j=0}^{i-1} \frac{x-a_j}{i+1-a_j} \prod_{j=i+1}^n \frac{x-a_j}{i+1-a_j}$$where $a_m$ is $m$ if $m<k$ and $m+1$ if $m\geq k$. If $i<k$ then \begin{align*} & \prod_{j=0}^{i-1} \frac{x-a_j}{i-a_j}=\prod_{j=0}^{i-1} \frac{x-j}{i-j}=\binom{x}i \\ & \prod_{j=i+1}^n \frac{x-a_j}{i -a_j}=\frac{i-k}{x-k} \frac{(x-1-i)!}{(-1)^{x-1-i}(x-1-i)!} =(-1)^{x-i} \frac{k-i}{x-k}\end{align*}We get a similar thing for $i\geq k$. We find \begin{align*} P(n+2) &= \sum_{i=0}^{k-1} 2^i (-1)^{n-i} \frac{k-i}{n+2-k} \binom{n+2}i+ \sum_{i=k}^{n} 2^{i+1} (-1)^{n-i-1} \frac{k-i-1}{n+2-k} \binom{n+2}{i+1} \\ &= \sum_{i=0}^{n+1} 2^i (-1)^{n-i} \frac{k-i}{n+2-k} \binom{n+2}i- 2^k (-1)^{n-k} \frac{k-k}{n+2-k} \binom{n+2}k \\ &= \frac{1}{n+2-k} \sum_{i=0}^{n+1} 2^i (-1)^{n-i} (k-i)\binom{n+2}i\end{align*}we can find, using the same calculus way as before, this is equal to $2^{n+2}+\frac{k-2n-4}{n+2-k}$. And, $$2^{n+2}+\frac{k-2n-4}{n+2-k}=2^{n+2}-n-3\implies k=n+1$$contradiction.