You actually had just about everything right, except that you skipped an important step: your normal vector to the surface $ \ \vec{ds} \ $ is correct, but you need to integrate its length over the surface of the cone nappe in order to obtain the surface area.
I'll generalize the problem a little, since the choice of proportions for the cone hides one of the factors in the surface area result. For a cone nappe with a height $ \ h \ $ and a "base radius" $ \ r \ $ , we can use similar triangles to find the parametrization (using your notation)
$$ x \ = \ \left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ y \ = \ \left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ z \ = \ u \ \ , $$
with the domain $ \ 0 \ \le \ u \ \le \ h \ , \ 0 \ \le \ p \ < \ 2 \pi \ $ . An "upward" normal vector is then given by
$$ \vec{R_u} \ \times \ \vec{R_p} \ \ " = " \ \ \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \left( \frac{r}{h} \right) \cos \ p&\left( \frac{r}{h} \right) \sin \ p\quad&1\\ -\left( \frac{r}{h} \right) u \ \sin \ p&\left( \frac{r}{h} \right) u \ \cos \ p\quad&0\end{array}\right| $$
$$ = \ \langle \ -\left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ -\left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ \left( \frac{r}{h} \right)^2 u \ \rangle \ \ . $$
So, up to this point, your procedure is fine. What is needed now is the "norm" of this vector:
$$ \| \ \vec{R_u} \ \times \ \vec{R_p} \ \| \ \ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ \cos^2 \ p \ + \ \left( \frac{r}{h} \right)^2 u^2 \ \sin^2 \ p \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ \ . $$
$$ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ . $$
It is the "magnitude" of the infinitesimal patches associated with the normal vectors that we wish to integrate over the domain of the parameters. Thus,
$$ S \ \ = \ \ \int_0^{2 \pi} \int_0^h \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ du \ dp $$
$$ = \ \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \int_0^{2 \pi} dp \ \int_0^h \ u \ \ du $$
$$ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ 2 \pi \ \cdot \ \left(\frac{1}{2}u^2 \right) \vert_0^h \ \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ \pi \ h^2 $$
$$ = \ \pi \ h \ \cdot \ \left(\frac{r}{h} \right) \ \cdot \ h \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ = \ \pi \ r \ \sqrt{ r^2 \ + \ h^2 } \ \ , $$
or $ \ \pi \ $ times the "base radius" times the "slant height" of the cone nappe, as the surface area is frequently expressed. In your use of the "standard cone", for which $ \ r \ = \ h \ $ , this formula gives us $ \ S \ = \ \pi \ \sqrt{2} \ h^2 \ $ , as you will find for your calculations, with the restoration of the omitted step.
It is very nice you be told that your answer is not correct.
$\sin(x/4)$ is zero if $x = 4 \pi$ (this is your 12.566371 which is just an approximate value). Then $a = 4 \pi$. If you plug this number in your integral (we are still waiting for it), the exact result is $64 (\pi^2 - 4)$, the approximate value of which being
$$375.65468166971895160540742399207367266\ldots$$
So, your value of $375.655$ is wrong.
Best Answer
$(1)\;$ First question: $$y = (4 + x)^{1/2}, \;\; y = (4 - x)^{1/2}, \;\; y = 0 \;\text{(x-axis)}\;\;-4\leq x \leq 4$$
Graphing always helps. See the picture below, compliments of Wolfram Alpha. By setting the equations equal to one another, we can find (and see) that the point of intersection of the first two equations is at $(0, 2)$.
Note that for $x \in [-4, 0]$, we need the area bounded by $y = \sqrt{x + 4}$ and $y = 0$. For $x \in [0, 4]$, we want to compute the area bounded by $y = \sqrt{4 - x}\,$ and $\,y = 0.\;$.
So to compute the area bounded by the respective curves and the $x$-axis $\,(y = 0),\;$ we will compute the sum of two integrals:
$$\textrm{AREA }\; = \int_{-4}^0 \sqrt{x + 4} \, dx + \int_0^4 \sqrt{4 - x} \; dx\tag{1}$$
$(2)\;$ Second problem: $$y = \sin x,\quad y = \cos x, \quad \frac{\pi}{4} \leq x \leq \frac{5 \pi}4$$
Try to find the points of intersection for your second problem by equating $\sin x = \cos x$. Graph the curves so you can see which function is the upper bound of the area to be computed, on the given interval, and which function is the lower bound.
We see $\sin x > \cos x$ over the given interval, and that the $x$-coordinates of the two points of intersection of the curves are precisely the bounds for integrating: from $\;\frac{\pi}{4}\;$ to $\;\frac{5 \pi}4.\;$ In the integrand, we subtract $\cos x\,$ (lower bound) from $\,\sin x\,$ (upper bound).
$$\textrm{AREA } \; = \int_{\pi/4}^{5\pi/4}\left( \sin x - \cos x \right)\; dx\tag{2}$$
I'll let you take the evaluation of the integrals from here!