jmc

Squaring the equation $a+b+c=1,$ we get $$a^2+b^2+c^2+2ab+2ac+2bc=1.$$Since $a^2+b^2+c^2=2,$ $2ab+2ac+2bc=-1,$ so $$ab+ac+bc=-\frac{1}{2}.$$Cubing the equation $a+b+c=1,$ we get $$(a^3+b^3+c^3)+3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)+6abc=1.$$Since $a^3+b^3+c^3=3,$ $$3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)+6abc=-2.(\ast )$$If we multiply the equations $a+b+c=1$ and $a^2+b^2+c^2=2,$ we get $$(a^3+b^3+c^3)+(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)=2.$$Then $$a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2=-1.$$Then from equation $(\ast ),$ $$-3+6abc=-2,$$so $abc=\frac{1}{6}.$

By Vieta's formulas, $a,$ $b,$ $c,$ are the roots of the equation $x^3-x^2-\frac{1}{2}x-\frac{1}{6}=0.$ Hence, $$\begin{array}{rl}{a}^3-a^2-\frac{1}{2}a-\frac{1}{6}& =0,\\ b^3-b^2-\frac{1}{2}b-\frac{1}{6}& =0,\\ c^3-c^2-\frac{1}{2}c-\frac{1}{6}& =0.\end{array}$$Multiplying these equations by $a,$ $b,$ $c,$ respectively, we get $$\begin{array}{rl}{a}^4-a^3-\frac{1}{2}{a}^2-\frac{1}{6}a& =0,\\ b^4-b^3-\frac{1}{2}{b}^2-\frac{1}{6}b& =0,\\ c^4-c^3-\frac{1}{2}{c}^2-\frac{1}{6}c& =0.\end{array}$$Adding these equations, we get $$(a^4+b^4+c^4)-(a^3+b^3+c^3)-\frac{1}{2}(a^2+b^2+c^2)-\frac{1}{6}(a+b+c)=0,$$so $$a^4+b^4+c^4=(a^3+b^3+c^3)+\frac{1}{2}(a^2+b^2+c^2)+\frac{1}{6}(a+b+c)=3+\frac{1}{2}\cdot 2+\frac{1}{6}\cdot 1=\boxed{\frac{25}{6}}.$$