Ok so it looks like I misjudged this and the community seem happy to have the question here, at least at present. So I figured I'd pass on the comments which Serre sent the LMS.
p.135, part b) of Lemma 4. Replace $H^q(H,M)$ by $H^q(K/H,M^H)$ and replace $\hat{H}^0(H,M)$ by $\hat{H}^0(K/H,M^H)$.
p.135, line 6 from bottom. Replace "to $G/H$" by "to $H$".
p.145. In Prop.6, replace "of Proposition 3" by "of Proposition 5".
p.225, line 11 (second line after Th.7). After "can be taken to be rational
integers" add the following parenthesis : (provided one does not insist that the
extensions $K/\Omega_{\ell}$ be cyclic).
Two which are for food rather than cash:
Let $f = t^{2d} + f_1 t^{2d-1} + f_2 t^{2d-2}+ \cdots f_d t^d + \cdots+ f_2 t^2 +f_1 t + 1$ be a palindromic polynomial, so the roots of $f$ are of the form $\lambda_1$, $\lambda_2$, ..., $\lambda_d$, $\lambda_1^{-1}$, $\lambda_2^{-1}$, ..., $\lambda_d^{-1}$. Set $r_k = \prod_{j=1}^d (\lambda_j^k-1)(\lambda_j^{-k} -1)$.
Conjecture: The coefficients of $f$ are uniquely determined by the values of $r_1$, $r_2$, ... $r_{d+1}$.
Motivation: When computing the zeta function of a genus $d$ curve over $\mathbb{F}_q$, the numerator is essentially of the form $f$. (More precisely, it is of the form $q^d f(t/\sqrt{q})$ for $f$ of this form.) Certain algorithms proceed by computing the $r_k$ and recovering the coefficients of $f$ from them. Note that you have to recover $d$ numbers, so you need at least $r_1$ through $r_d$; it is known that you need at least one more and the conjecture is that exactly one more is enough.
Reward: Sturmfels and Zworski will buy you dinner at Chez Panisse if you solve it.
Consider the following probabilistic model: We choose an infinite string, call it $\mathcal{A}$, of $A$'s, $C$'s, $G$'s and $T$'s. Each letter of the string is chosen independently at random, with probabilities $p_A$, $p_C$, $p_G$ and $p_T$.
Next, we copy the string $\mathcal{A}$ to form a new string $\mathcal{D}_1$. In the copying process, for each pair $(X, Y)$ of symbols in $\{ A, C, G, T \}$, there is some probability $p_1(X \to Y)$ that we will miscopy an $X$ as a $Y$. (The $16$ probabilities stay constant for the entire copying procedure.)
We repeat the procedure to form two more strings $\mathcal{D}_2$ and $\mathcal{D}_3$, using new probability matrices $p_2(X \to Y)$ and $p_3(X \to Y)$.
We then forget the ancestral string $\mathcal{A}$ and measure the $64$ frequencies with which the various possible joint distributions of $\{ A, C, G, T \}$ occur in the descendant strings $(\mathcal{D}_1, \mathcal{D}_2, \mathcal{D}_3)$.
Our procedure depended on $4+3 \times 16$ inputs: the $(p_A, p_C, p_G, p_T)$ and the $p_i(X \to Y)$. When you remember that probabilities should add up to $1$, there are actually only $39$ independent parameters here, and we are getting $63$ measurements (one less than $64$ because probabilities add up to $1$). So the set of possible outputs is a semialgeraic set of codimension $24$.
Conjecture: Elizabeth Allman has a conjectured list of generators for the Zariski closure of the set of possible measurements.
Motivation: Obviously, this is a model of evolution, and one which (some) biologists actually use. Allman and Rhodes have shown that, if you know generators for the ideal for this particular case, then they can tell you generators for every possible evolutionary history. (More descendants, known sequence of branching, etc.) There are techniques in statistics where knowing this Zariski closure would be helpful progress.
Reward: Elizabeth Allman will personally catch, clean, smoke and ship an Alaskan Salmon to you if you find the generators. (Or serve it to you fresh, if you visit her in Alaska.)
Best Answer
There is the famous Unsolved Problems in Number Theory by Richard Guy, 3rd Edition, 2004. PDF available at the Springer link.