First problem: I have no idea where to start with the problem, so little a direction on where to start would be very appreciated.
my handwritten work for second problem (sorry it's sideways)
second problem: I'm stuck on the last step of verifying my identity, I honestly have no idea how to move forward so a walkthrough of the steps would be appreciated.
On
For the first problem, stare at the Fundamental Theorem of Calculus until you see why what you were given shows that the theorem applies. Then we have $$ 21=\int_e^\pi f'(x) \;dx = f(x)\bigg|_e^\pi = f(\pi)-f(e)=f(\pi)-13 $$ But then $f(\pi)-13=21$ so that $f(\pi)=34$.
For the second problem, $$ \begin{split} 2 \int_a^x f(t) \; dt = 2 \sin x -1 \\ \int_a^x f(t) \; dt = \sin x - \frac{1}{2} \end{split} $$ Differentiate and use the Fundamental Theoem of Calculus: $$ \begin{split} \dfrac{d}{dx} \int_a^x f(t) \; dt &= \dfrac{d}{dx} \left[\sin x - \frac{1}{2}\right] \\ f(x) &= \cos x \end{split} $$ Now it is just a matter of finding the value $a$. $$ \int_a^x f(t) \; dt = \int_a^x \cos t \; dt = \sin t \,\bigg|_a^x= \sin x - \sin a $$ But we need this to be $\sin x - \frac{1}{2}$. Then $\sin x - \sin a = \sin x - \frac{1}{2}$. This means $\sin a = \frac{1}{2}$. But then we need $a= \arcsin \frac{1}{2}= \frac{\pi}{6}$. In short, you had all the right ideas!
First problem: think Fundamental Theorem of Calculus.
Second problem: You're almost there. Evaluate the integral: $\int_{\frac\pi6}^x \cos t dt = \sin t|_\frac\pi6^x = \sin x - \sin\frac\pi6$