(a) The product rule gives \[\frac{dy}{dx}=e^{2x}(7\cos x-\sin x).\] Hence \(\tan x=7\). The one solution in the interval is \(\boxed{x=\tan^{-1}7\approx1.43}\), in radians.

(b) \[\frac{d^2y}{dx^2}=e^{2x}(13\cos x-9\sin x).\] At the stationary point, \(\sin x=7\cos x\), so this equals \(-50e^{2x}\cos x<0\). Therefore it is a maximum.

(a) By the product rule, \[\frac{dy}{dx}=e^{1-2x}-2xe^{1-2x}=e^{1-2x}(1-2x).\] Since the exponential never vanishes, the stationary point occurs at \(x=1/2\), with \(y=(1/2)e^0=1/2\). Thus \[\boxed{(1/2,1/2)}.\]

(b) Differentiating again gives \[\frac{d^2y}{dx^2}=4(x-1)e^{1-2x}.\] At x=1/2 this is −2, so the point is a \(\boxed{\text{maximum}}\).

By the product rule, \[y^{\prime}=-5e^{-5x}\tan^2x+2e^{-5x}\tan x\sec^2x\] \[=e^{-5x}\tan x(2\sec^2x-5\tan x).\] The exponential does not vanish. The factor tan x=0 gives x=0. For the remaining points, put t=tan x and use sec²x=1+t²: \[2(1+t^2)-5t=0,\qquad(2t-1)(t-2)=0.\] Thus t=1/2 or 2. Tangent is one-to-one on the given interval, so \[\boxed{x=0,\quad0.464,\quad1.107\text{ radians}}.\] All three roots lie in the specified domain.

By the product rule, \[\frac{dy}{dx}=-\sin x\sin2x+2\cos x\cos2x.\] Using double-angle identities and the positive factor \(\cos x\), a zero derivative requires \[-\sin^2x+\cos^2x-\sin^2x=0\quad\Longrightarrow\quad\sin^2x=1/3.\] Therefore \(\boxed{x=\sin^{-1}(1/\sqrt3)\approx0.615}\), in radians. Sine increases strictly on the stated interval, so there is one stationary point.

(a) By the product and chain rules, \[\boxed{\frac{dy}{dx}=\tan^{-1}(x/2)+\frac{2x}{x^2+4}}.\]

(b) At x=2, \(y=\pi/2\), and the gradient is \(\pi/4+1/2\). Thus \[y-\pi/2=(\pi/4+1/2)(x-2).\] Setting x=0 gives \(\boxed{p=-1}\).

By the product and chain rules, \[\frac{dy}{dx}=e^{\sin2x}\bigl(2\cos2x\cos4x-4\sin4x\bigr)=2e^{\sin2x}\cos2x\bigl(\cos4x-4\sin2x\bigr).\] For an interior stationary point, \(\cos2x>0\). Put \(s=\sin2x\) and use \(\cos4x=1-2s^2\): \[1-2s^2-4s=0\Rightarrow s=\frac{-2\pm\sqrt6}{2}.\] Only \((\sqrt6-2)/2\) lies in \((0,1)\). The derivative changes from positive to negative, so this gives \(M\): \[x=\frac12\arcsin\left(\frac{\sqrt6-2}{2}\right)=0.113\text{ radians}\quad(3\text{ s.f.}).\]

Since \(x>0\), \(\ln(x^4)=4\ln x\). Thus \[y=12x^3\ln x,\qquad y\prime=36x^2\ln x+12x^2=12x^2(3\ln x+1).\] As \(x\ne0\), stationarity gives \(\ln x=-1/3\), so \(x=e^{-1/3}\). Substituting, \[y=12e^{-1}\left(-\frac13\right)=-\frac4e.\] The stationary point is \[\left(e^{-1/3},-\frac4e\right).\]

\(y^\prime=2e^{2x}(\sin2x+\cos2x)\). Setting it to zero gives \(\tan2x=-1\); since \(0\le2x\le\pi\), \(2x=3\pi/4\). Thus \[\left(x,y\right)=\left(\frac{3\pi}{8},\frac{\sqrt2}{2}e^{3\pi/4}\right).\]

By the product rule, \[\frac{dy}{dx}=5e^{x/2}+\frac52xe^{x/2}=\frac52e^{x/2}(x+2).\] Since \(e^{x/2}>0\), the derivative is zero only when \(x=-2\). Then \(y=5(-2)e^{-1}=-10/e\). The stationary point is \(\boxed{(-2,-10/e)}\).