It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the frattini subgroup of $G$ is trivial and the maximal subgroups are (that can be check by GAP):  $M_1=\langle(6,7),(1,2,3,4,5)\rangle$,  $M_2=\langle (2,5)(3,4),(1,2,3,4,5)\rangle$,  $M_3=\langle(2,5)(3,4)(6,7),(1,2,3,4,5)\rangle$,  $M_4=\langle(6,7),(2,5)(3,4)\rangle$,  $M_5=\langle(6,7),(1,4)(2,3)\rangle$,  $M_6=\langle(6,7),(1,2)(3,5)\rangle$,  $M_7=\langle(6,7),(1,5)(2,4)\rangle$, and  $M_8=\langle(6,7),(1,3)(4,5)\rangle$. Now consider $H_1=\langle(6,7)\rangle$ and $H_2=\langle(1,2,3,4,5)\rangle$.

It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the Frattini subgroup of $G$ is trivial, and the maximal subgroups of $G$ are (this can be checked by GAP): 

$M_1=\langle(6,7),(1,2,3,4,5)\rangle$, 

$M_2=\langle (2,5)(3,4),(1,2,3,4,5)\rangle$, 

$M_3=\langle(2,5)(3,4)(6,7),(1,2,3,4,5)\rangle$, 

$M_4=\langle(6,7),(2,5)(3,4)\rangle$, 

$M_5=\langle(6,7),(1,4)(2,3)\rangle$, 

$M_6=\langle(6,7),(1,2)(3,5)\rangle$, 

$M_7=\langle(6,7),(1,5)(2,4)\rangle$, and 

$M_8=\langle(6,7),(1,3)(4,5)\rangle$. 

Now consider $H_1=\langle(6,7)\rangle$ and $H_2=\langle(1,2,3,4,5)\rangle$.

It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the frattini subgroup of $G$ is trivial and the maximal subgroups are (that can be check by GAP): $M_1=<(6,7),(1,2,3,4,5)>$, $M_2=<(2,5)(3,4),(1,2,3,4,5)>$, $M_3=<(2,5)(3,4)(6,7),(1,2,3,4,5)>$, $M_4=<(6,7),(2,5)(3,4)>$, $M_5=<(6,7),(1,4)(2,3)>$, $M_6=<(6,7),(1,2)(3,5)>$, $M_7=<(6,7),(1,5)(2,4)>$, and $M_8=<(6,7),(1,3)(4,5)>$. Now consider $H_1=<(6,7)>$ and $H_2=<(1,2,3,4,5)>$.

It seems that there are some examples. Let $G=D_{10}\times S_2$. Then the frattini subgroup of $G$ is trivial and the maximal subgroups are (that can be check by GAP): $M_1=\langle(6,7),(1,2,3,4,5)\rangle$, $M_2=\langle (2,5)(3,4),(1,2,3,4,5)\rangle$, $M_3=\langle(2,5)(3,4)(6,7),(1,2,3,4,5)\rangle$, $M_4=\langle(6,7),(2,5)(3,4)\rangle$, $M_5=\langle(6,7),(1,4)(2,3)\rangle$, $M_6=\langle(6,7),(1,2)(3,5)\rangle$, $M_7=\langle(6,7),(1,5)(2,4)\rangle$, and $M_8=\langle(6,7),(1,3)(4,5)\rangle$. Now consider $H_1=\langle(6,7)\rangle$ and $H_2=\langle(1,2,3,4,5)\rangle$.