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(a) If the integrand had $x$'s instead of $e^x$'s it would be a rational function, ripe for the application of partial fractions. So let's start by making the substitution $u=e^x$, $\mathrm{d}u=e^x\,\mathrm{d}x$:

$$\begin{aligned} \int \frac{e^x}{(e^x+1)(e^x-3)} \mathrm{d}x = \int \frac{\mathrm{d}u}{(u+1)(u-3)} \end{aligned}$$

Now, we follow the partial fractions protocol, starting with expressing

$$\begin{aligned} \frac{1}{(u+1)(u-3)} = \frac{A}{u+1} + \frac{B}{u-3} \end{aligned}$$

To find $A$ and $B$, the sneaky way, we cross multiply by the denominator

$$\begin{aligned} 1 = A(u-3) + B(u+1) \end{aligned}$$

and find $A$ and $B$ by evaluating at $u=-1$ and $u=3$, respectively.

$$\begin{aligned} 1 = A(-1-3) + B(-1+1) &\iff A=-\frac{1}{4} \\ 1 = A(3-3) + B(3+1) &\iff B=\frac{1}{4} \\ \end{aligned}$$

Finally, we can do the integral:

$$\begin{aligned} \int \frac{e^x}{(e^x+1)(e^x-3)} \mathrm{d}x &= \int \frac{\mathrm{d}u}{(u+1)(u-3)} = \int \Big(\frac{ -1/4}{u+1} + \frac{ 1/4}{u-3}\Big)\mathrm{d}u \\ &=-\frac{1}{4}\log|u+1| +\frac{1}{4}\log|u-3| + C \\ &=-\frac{1}{4}\log|e^x+1| +\frac{1}{4}\log|e^x-3| + C \end{aligned}$$

(b) The argument of the square root is

$$\begin{aligned} 12+4x-x^2 =12-(x-2)^2 +4 =16-(x-2)^2 \end{aligned}$$

Hmmm. The numerator is $x^2-4x+4=(x-2)^2$. So let's make the integral look somewhat simpler by substituting $u=x-2$, $\mathrm{d}u=\mathrm{d}x$. When $x=2$ we have $u=0$, and when $x=4$ we have $u=2$, so:

$$\begin{aligned} \int_{x=2}^{x=4} \frac{x^2-4x+4}{\sqrt{12+4x-x^2}} \mathrm{d}x =\int_{u=0}^{u=2} \frac{u^2}{\sqrt{16-u^2}} \mathrm{d}u \end{aligned}$$

This is perfect for the trig substitution $u=4\sin\theta$, $\mathrm{d}u = 4\cos(\theta)\,\mathrm{d}\theta$. When $u=0$ we have $4\sin\theta = 0$ and hence $\theta=0$. When $u=2$ we have $4\sin\theta = 2$ and hence $\theta=\frac{\pi}{6}$. So

$$\begin{aligned} \int_{u=0}^{u=2} \frac{u^2}{\sqrt{16-u^2}} \mathrm{d}u &=\int_{\theta=0}^{\theta=\pi/6} \frac{16\sin^2\theta}{\sqrt{16-16\sin^2\theta}} 4\cos\theta\mathrm{d}\theta \\ &=16\int_{0}^{\pi/6} \sin^2\theta\ \mathrm{d}\theta \\ &=8\int_{0}^{\pi/6} \big(1-\cos(2\theta)\big)\ \mathrm{d}\theta \\ &=8 \bigg[\theta-\frac{1}{2}\sin(2\theta)\bigg]_{0}^{\pi/6} =8 \bigg[\frac{\pi}{6}-\frac{1}{2}\cdot \frac{\sqrt{3}}{2} \bigg] \\ & = \frac{4\pi}{3}-2\sqrt{3} \end{aligned}$$