Consider the SDE $$ d X_t = a X_t d t + b X_t d W_t, \ X_0 = x_0, $$ $X_t\in [0,\infty)$, where $a$, $b$, $x_0$ are all positive constants.
(a) I want to show that $$ \frac{d}{d t}\mathbb{E}[ X_t] = a\mathbb{E}[X_t], \\ \frac{d}{d t}\mathbb{E}\big[X^2_t\big] = \big(2a + b^2\big)\mathbb{E}\big[X^2_t\big], \\ \text{Var}[X_t] = x_0^2 e^{2at}\left(e^{b^2t}-1\right). $$ and $$ \mathbb{E}\big[ X_s X_{\tau}\big], $$ where $0\le s\le \tau$. Suppose $X_{t}=g(t,W_{t})$, by Ito's formula, we have $$ X_{T}-X_{0}=\int_{0}^{T}g_{x}(t,x)\, d W_{t}+\int_{0}^{T}g_{t}(t,x)\, d t+\frac {1}{2}\int_{0}^{T}g_{xx}(t,x)\, d t. $$ To solve this SDE, we need to find a function $g$ satisfying: $$ \partial_{x}g(t,x)=bx\quad\text{and}\quad \frac {1}{2}\partial_{xx}g(t,x)+\partial_{t}g(t,x)=ax. $$ By solving the above two equations, we get $g(t,x)=x_{0}\text{exp}\left(\left(a-\frac {1}{2}b^2\right)t+bx\right)$. Therefore, we conclude that $X_{t}=x_{0}\text{exp}\left(\left(a-\frac {1}{2}b^2\right)t+bW_{t}\right)$ solves this SDE. Since $W_{t}\sim$N($0,t$), we have $$ \mathbb {E}\left[X_{t}\right]=x_{0}e^{\left(a-\frac {1}{2}b^2\right)t}. $$ However, $$ \frac{d}{d t}\mathbb{E}[ X_t] =\left(a-\frac {1}{2}b^{2}\right)x_{0}e^{\left(a-\frac {1}{2}b^2\right)t}\neq a\mathbb{E}[X_t]. $$ I wonder what's the issue here.
The mistake lies in the penultimate step. You have $X_t =x_0 e^{\left(a-\frac{b^2}{2}\right)t +b W_t}$. It follows then that \begin{align} \mathbb{E}(X_t) = x_0e^{\left(a-\frac{b^2}{2}\right)t} \int_{\mathbb{R}} e^{bx}\mathrm{d}\mu_t(x) \, , \end{align} where $\mu_t=\mathcal{N}(0,t)$. We can simplify this as follows \begin{align} \mathbb{E}(X_t) =& x_0e^{\left(a-\frac{b^2}{2}\right)t} \int_{\mathbb{R}} e^{\frac{b^2}{2}t}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-bt)^2}{2 t}} \, \mathrm{d}x\, , \\ =&x_0 e^{at} \, . \end{align} Differentiate and you will have your answer.
Edit: For the second part set $g(t,x)=x^2$. Then, by Itô's formula, we have \begin{align} d g(t,X_t) = a g'(t,X_t) X_t\, \mathrm{d}t + \frac{b^2}{2}g''(t,X_t) X_t^2 \, \mathrm{d}t + b g'(t,X_t)X_t \, dW_t \, . \end{align} Plugging in the values of $g,g',g''$, we obtain \begin{align} dX_t^2 = 2a X_t^2 \, \mathrm{d}t + b^2 X_t^2 \, \mathrm{d}t + 2 b X_t^2 \, dW_t \, . \end{align} Take expectation and you will have your result.