How to show $M_n = X_n^2-n$ is a martingale?

For $s\le n$, $$\mathbb{E}(M_{n+1}\mid \mathcal{F}_s)=\mathbb{E}((X_n+Y_{n+1})^2-(n+1)\mid \mathcal{F}_s)\\=X_n^2+2X_n\mathbb{E}(Y_{n+1})+\mathbb{E}(Y_{n+1}^2)-(n+1)\\=X_n^2-n=M_n$$ the steps have used the fact that $Y_n$ is an i.i.d. $\mathcal{N}(0,1)$ sequence. The possible mistake that you are making in your approach is the one rightly pointed out by @saz in the comments.

For part (B) $$\mathbb{E}M_n=\mathbb{E}X_n^2-n$$ Note that $X_n=\sum_{i=1}^n Y_i+X_0\implies \mathbb{E}X_n^2=n+X_0^2\implies \mathbb{E}M_n=X_0^2$

Let's look at $\mathbb{E}[X_{n+1}^2-(n+1)|F_n]$ (take $s=n$). We need show it equals to $X_n^2-n$. Since $X_{n+1}=X_n+ Y_{n+1}$ It should be $$ \mathbb{E}[X_{n}^2|F_n]+2\mathbb{E}[X_{n}Y_{n+1}|F_n]+\mathbb{E}[Y_{n+1}^2|F_n]-\mathbb{E}[n+1|F_n] $$, Since $X_{n}$ is measurable w.r.t. $F_n$, and $Y_{n+1}$ is independent of $F_n$ (for which we can "take out what is known, and by the assumption $Y_n$ is $\mathcal{N}(0,1)$ distributed, we get above equals

$$ X_{n}^2+2X_{n}\mathbb{E}[Y_{n+1}|F_n]+\mathbb{E}[Y_{n+1}^2]-(n+1), $$

$$ =X_{n}^2+2X_{n}\mathbb{E}[Y_{n+1}]+\mathbb{E}[Y_{n+1}^2]-(n+1). $$ Since $\mathbb{E}[Y_{n+1}]=0$, it's easy to see $\mathbb{E}[Y_{n+1}^2]=1$. Hence we reach $$ =X_{n}^2-n. $$