If $f$ derivable on $[a,b]$ does $\int_a^t f'(x)dx=f(t)-f(a)$ true?

Q1: No, because $f'$ need not be Riemann integrable. For example $f'$ need not be bounded.

Q2: Yes. This is in fact very simple - the elegance of the proof seems to me to be a good reason for the Riemann integral to be defined exactly the way it is. Say $a=t_0<\dots<t_n=b$. Apply the Mean Value Theorem to each subinterval: $$f(b)-f(a)=\sum(f(t_{j+1})-f(t_j))=\sum f'(\xi_j)(t_{j+1}-t_j),$$precisely a Riemann sum for $\int_a^b f'(t)\,dt$.

Q3: No, $f'$ need not be Lebesgue integrable. For example if $a=-1$, $b=1$ and $$f(t)=\begin{cases}t^2\sin(1/t^{100}),&(t\ne0), \\0,&(t=0).\end{cases}$$

Q4: Pretty sure the answer is yes. Almost certain this is a theorem in Rudin - my copy of Real and Complex Analysis is missing. Edit: Thanks to @ChrisJanjigian for confirming that yes, it's Theorem 7.21. Or see here for a proof based on the Vitali-Caratheodory theorem.

Note this is assuming that $f$ is differentiable, which is to say that $f'$ exists at every point. There are certainly examples where $f$ is differentiable almost everywhere, $f'\in L^1$, but $f$ is not absolutely continuous; for example the Cantor function satisfies $f'=0$ almost everywhere.