It should be \(30^{\circ}\) and \(105^{\circ}\). See Indian RMO 2000~2005.

I'm not so good at Asymptote, but still... Sharygin Finals 2019 Grade 8 P5 wrote: A triangle having one angle equal to $45^{\circ}$ is drawn on the chequered paper. Find the values of its remaining angles [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.419734379119163, xmax = 22.870506921812265, ymin = -9.695220833689252, ymax = 7.642650575679416; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-5.90870452245749,3.6364773877125542)--(16.498428913318573,-2.255110794329651)--(6.140377906573217,-8.300219914547753)--cycle, linewidth(2)); draw(arc((16.498428913318573,-2.255110794329651),0.769432163137658,165.26845682713358,210.26845682713358)--(16.498428913318573,-2.255110794329651)--cycle, linewidth(1.6) + rvwvcq); /* draw figures */ draw((-5.90870452245749,3.6364773877125542)--(-5.88060824076489,-2.3599675451102), linewidth(0.8) + dtsfsf); draw((-5.88060824076489,-2.3599675451102)--(-5.852511959072291,-8.356412477932954), linewidth(0.8) + dtsfsf); draw((-5.90870452245749,3.6364773877125542)--(16.498428913318573,-2.255110794329651), linewidth(2)); draw((16.498428913318573,-2.255110794329651)--(6.140377906573217,-8.300219914547753), linewidth(2)); draw((6.140377906573217,-8.300219914547753)--(-5.90870452245749,3.6364773877125542), linewidth(2)); draw((-5.852511959072291,-8.356412477932954)--(18.13326777221873,-8.244027351162556), linewidth(0.8) + dtsfsf); draw((18.13326777221873,-8.244027351162556)--(18.077075208833527,3.7488625144829526), linewidth(0.8) + dtsfsf); draw((0.08774041036526398,3.664573669405154)--(0.14393297375046318,-8.328316196240356), linewidth(0.4) + dtsfsf); draw((6.084185343188018,3.6926699510977534)--(6.140377906573217,-8.300219914547753), linewidth(0.4) + dtsfsf); draw((-5.90870452245749,3.6364773877125542)--(18.077075208833527,3.7488625144829526), linewidth(0.8) + dtsfsf); draw((12.080630276010773,3.720766232790353)--(12.136822839395974,-8.272123632855156), linewidth(0.4) + dtsfsf); draw((-5.88060824076489,-2.3599675451102)--(18.10517149052613,-2.247582418339802), linewidth(0.4) + dtsfsf); /* dots and labels */ label("$45^{\circ}$", (16.022560669887106,-3.3602293571891613), NE * labelscalefactor,rvwvcq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

We use designation from picture. Since $\triangle ABO$ is right isosceles, $O$ is a circumcenter of $\triangle ABC$. From symmetry wrt $CM$ we obtain $|OC|=|BC|$, so $\triangle BCO$ is equilateral. Hence $\angle BAC=30^{\circ}$ and $\angle ABC=105^{\circ}$.