Statement 1 : ${\mathrm{H}}_3{\mathrm{BO}}_3$ is monobasic Lewis acid but salt ${\mathrm{Na}}_3{\mathrm{BO}}_3\mathrm{\hspace{0.17em}\hspace{0.17em}}$ exists.

Statement 2 : ${\mathrm{H}}_3{\mathrm{BO}}_3$ reacts with NaOH to give ${\mathrm{Na}}_3{\mathrm{BO}}_3.$

H₃BO₃ in it self does not contain any replaceable hydrogen atom

But on interaction with water a hydrogen ion is liberated by  accepting OH^- ion

Hence it is a monobasic acid 

Due to the presence of empty orbital on Boron atom , It  readily accepts the electron pair and acts as a Lewis acid

$\left(\mathrm{II}\right) \mathrm{B}{\left(\mathrm{OH}\right)}_3\mathrm{\hspace{0.17em}}+\hspace{0.17em}\hspace{0.17em}\mathrm{NaOH}\hspace{0.17em}\hspace{0.17em}\to \hspace{0.17em}\hspace{0.17em}\mathrm{Na}\mathrm{\hspace{0.17em}}{\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right]}^{-}$