Expand the square of the sum:

\((\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\).

Since \(\sin^2\alpha+\cos^2\alpha=1\) and \(2\sin\alpha\cos\alpha=\sin2\alpha\), we get

\((\sin\alpha+\cos\alpha)^2=1+\sin2\alpha\).

For \(\alpha=-15^\circ\):

\(1+\sin(-30^\circ)=1-\dfrac12=\dfrac12\).

Use the definition of tangent:

\(\tan^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}\).

Then

\(\tan^2\alpha\cdot2\cos^2\alpha=2\sin^2\alpha\).

Therefore, the whole expression equals

\(2-2\sin^2\alpha=2(1-\sin^2\alpha)=2\cos^2\alpha\).

For \(\alpha=-\dfrac{\pi}{4}\):

\(2\cos^2\left(-\dfrac{\pi}{4}\right)=2\cdot\dfrac12=1\).

Use the double-angle formula:

\(\sin4\alpha=2\sin2\alpha\cos2\alpha\).

Then

\(\dfrac{\sin4\alpha}{\cos2\alpha}=2\sin2\alpha\).

For \(\alpha=15^\circ\):

\(2\sin30^\circ=2\cdot\dfrac12=1\).

Transform the expression in parentheses:

\(\cos2\alpha+\sin^2\alpha=(\cos^2\alpha-\sin^2\alpha)+\sin^2\alpha=\cos^2\alpha\).

Then the whole expression becomes

\(6\cos^2\alpha\tan^2\alpha=6\cos^2\alpha\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha}=6\sin^2\alpha\).

For \(\alpha=\dfrac{3\pi}{4}\):

\(6\sin^2\dfrac{3\pi}{4}=6\cdot\dfrac12=3\).

Use the reduction formulas:

\(\tan(\pi-\alpha)=-\tan\alpha\), \(\sin\left(\dfrac{\pi}{2}-\alpha\right)=\cos\alpha\).

Then the expression becomes

\(4(-\tan\alpha)\cos\alpha=-4\dfrac{\sin\alpha}{\cos\alpha}\cos\alpha=-4\sin\alpha\).

For \(\alpha=\dfrac{5\pi}{6}\):

\(-4\sin\dfrac{5\pi}{6}=-4\cdot\dfrac12=-2\).