I share Artes's plea in the comments to mention how you obtained such magic constants like 1.2287280664334876
in your equations, as that might ease symbolic solutions. Nevertheless:
Method 1:
MeshFunctions
, once more, is usable for this kind of problem:
ParametricPlot[{1. (2.35 - 0.8603646545265691 Cos[t] Sin[t]^(1/4) +
1.2287280664334876 Sin[t]^(5/4)),
1. (0. - 1.2287280664334876 Cos[t] Sin[t]^(1/4) -
0.8603646545265691 Sin[t]^(5/4))}, {t, 0, 2 π},
Mesh -> {{0}},
MeshStyle -> Directive[RGBColor[0.7, 0, 0], AbsolutePointSize[5]],
MeshFunctions -> {Function[{x, y, t},
(-1 + x^2/5 + y^2/5)^3 - (x^2 y^2)/12]}]

Cases[Normal[%], Point[pt_] :> pt, ∞]
{{2.32563, -0.0349495}, {2.73159, 0.509025}, {2.94152, -1.31727}}
Method 2:
Use a Weierstrass substitution to help convert the parametric equation into an implicit Cartesian one:
tmp = Eliminate[MapAll[TrigExpand, Thread[{x, y} ==
{(2.35 - 0.8603646545265691 Cos[t] Sin[t]^(1/4) +
1.2287280664334876 Sin[t]^(5/4)),
(-1.2287280664334876 Cos[t] Sin[t]^(1/4) -
0.8603646545265691 Sin[t]^(5/4))} /. t -> 2 ArcTan[u]]], u];
(* alleviate sudden coefficient swell *)
tmp = Expand[Subtract @@ tmp];
tmp = Expand[tmp/Max[CoefficientList[tmp, {x, y}]]]
Now we can use NSolve[]
:
{x, y} /. NSolve[{tmp == 0, (-1 + x^2/5 + y^2/5)^3 - (x^2 y^2)/12 == 0}, {x, y}, Reals]
{{2.73164397095681, 0.509126233763010}, {2.42604085249915, 0.108583410750758},
{2.42599009519994, 0.108539249717106}, {2.32555967012562, -0.0349062696627526},
{2.32555782100012, -0.0349051879461614}, {2.94152191912221, -1.31727498432471}}
Show the intersection points:
Show[ContourPlot[(-1 + x^2/5 + y^2/5)^3 == (x^2 y^2)/12, {x, -4, 4}, {y, -4, 4}],
ParametricPlot[{1. (2.35 - 0.8603646545265691 Cos[t] Sin[t]^(1/4) +
1.2287280664334876 Sin[t]^(5/4)),
1. (0. - 1.2287280664334876 Cos[t] Sin[t]^(1/4) -
0.8603646545265691 Sin[t]^(5/4))}, {t, 0, 2 π}],
Epilog -> {Directive[RGBColor[0, 0.7, 0], AbsolutePointSize[6]], Point[%]}]

1.2287280664334876
and0.8603646545265691
. where do they come frome? You should know that providing exact values appears to be rewarding in Mathematica. $\endgroup$