does somebody have a solution?

Official solution. (a) The answer is yes. Andy can call the cards out in order starting with the Ace of Spades, two of Spades down to the King of Spades, followed by the Hearts, the Diamonds and the Clubs. We refer to this as one cycle. In each cycle, each card can move at most once since it is called exactly once, and at least one card must move. Andy then makes another $51$ cycles of calls. We claim that all moves are in the same direction, either all clockwise or all counter-clockwise. This is clear within each cycle. Consider the card $X$ which is the last to move in a cycle, and let $Y$ be the other card adjacent to the empty spot. Since $Y$ does not move after $X$ in this cycle, it must have been called before $X$. So in the next cycle, $Y$ will be called before $X$, and follows $X$ in the same direction. This justifies our claim. To go once around and return to its initial spot, a card must have moved $53$ times, and this is not possible since Andy makes only $52$ cycles of calls. If it is to be in its initial spot, it must not have moved at all. However, this is also impossible as otherwise at most 1 move could have been made, but in $52$ cycles, at least $52$ moves have been made. Therefore, after $52$ cycles of calls, every card is in a spot different from its initial one. (b) The answer is no. Construct a graph where each of the vertices represents one of the $52!$ permutations of the cards, with the first and the last adjacent to the empty spot. Two vertices are joined by an edge if and only if a call by Andy changes the two permutations to each other. Label the edge with the card called by Andy. In this graph, each vertex has degree $2$, and the graph is a union of disjoint cycles. Consider the cycle containing the vertex representing the initial permutation. For each vertex, let a person starts there. Whenever Andy makes a call, the person moves along an edge labelled with that card to an adjacent vertex if possible, and stays put otherwise. We call a vertex safe if and only if in the permutation it represents, the Queen of Spades is not adjacent to the empty spot. By shifting each card clockwise into the empty spot in turns, we will arrive at permutations represented by safe vertices as well as permutations represented by unsafe vertices. Note that after each call, there is still one person on each vertex. Thus no matter what sequence of calls Andy may employ, he cannot get everyone to a safe vertex. It follows that there is an initial permutation for which Andy's sequence will leave the Queen of Spades adjacent to the empty spot.

Here is another solution for (b), which is similar to the official but with a different approach (which i think is a bit more natural, although the idea of considering the graph is quite slick) Consider the set $X$ of possible configurations on the table, and note that $X$ is finite. Now, let $Y\subset X$ be the set of configurations for which the Queen of Spades is not adjacent to the empty spot. Enumerate the cards $1,2,\ldots$, and let $f_i\colon X\to X$ be the mapping from calling out card $i$. We wish to find $g = f_{i_1}\circ f_{i_2}\circ \cdots f_{i_k}$ so that $g(X) = Y$. The key observation is: \[ f_i\circ f_i = \mathrm{id} \]which is obvious, but a useful corollary is that $f_i$ is bijective, hence any such $g$ must be bijective, as it is the composition of bijections. Thus $g(X) = X$, so Andy must fail. $\blacksquare$