(a) Write t=tan θ. The addition and double-angle formulae give \[\tan4\theta=\frac{2\tan2\theta}{1-\tan^22\theta}=\frac{4t(1-t^2)}{1-6t^2+t^4}.\] For admissible solutions in the interval in part (b), t≠0; θ=90° is excluded because tan θ is undefined. Equating to t/2 and cancelling t gives \[8(1-t^2)=1-6t^2+t^4,\qquad\boxed{t^4+2t^2-7=0}.\] The intermediate cases θ=45° and 135°, where tan 2θ is undefined, do not solve the original equation.

(b) Put v=t²≥0. Then \[v^2+2v-7=0,\qquad v=-1\pm2\sqrt2.\] Only v=−1+2√2 is nonnegative, so \[\tan\theta=\pm\sqrt{-1+2\sqrt2}.\] Thus \[\boxed{\theta=53.5^\circ,\quad126.5^\circ}.\] Both satisfy the original equation; its tangent denominators are nonzero at these roots.